Utilisateur:Darkpatric/Brouillon

En mathématiques, les coefficients binomiaux de Gauss (également appelés coefficients de Gauss, polynômes de Gauss ou coefficients q- binomiaux ) sont des q-analogues des coefficients binomiaux . Le coefficient binomial de gauss, noté Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math> \binom nk_q} </mo></mrow><mfrac linethickness="0"><mi> ${\binom {n}{k}}_{q}$ </mi><mi> ${\binom {n}{k}}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${\binom {n}{k}}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${\binom {n}{k}}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${\binom {n}{k}}_{q}$ ${\binom {n}{k}}_{q}$ </img> ou Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mo> <math>\begin{bmatrix}n\\ k\end{bmatrix}_q} </mo><mtable rowspacing="4pt"><mtr><mtd><mi> ${\begin{bmatrix}n\\k\end{bmatrix}}_{q}$ </mi></mtd></mtr><mtr><mtd><mi> ${\begin{bmatrix}n\\k\end{bmatrix}}_{q}$ </mi></mtd></mtr></mtable><mo> ${\begin{bmatrix}n\\k\end{bmatrix}}_{q}$ </mo></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${\begin{bmatrix}n\\k\end{bmatrix}}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${\begin{bmatrix}n\\k\end{bmatrix}}_{q}$ ${\begin{bmatrix}n\\k\end{bmatrix}}_{q}$ </img> , est une fonction polynomiale en q avec des coefficients entiers qui, lorsque la valeur de q est fixé à une puissance d'un nombre premier, compte le nombre de sous-espaces de dimension k dans un espace vectoriel de dimension n sur un corps fini à q éléments.

Définition

Les coefficients binomiaux de Gauss sont définis par

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q = \begin{cases} \frac{(1-q^m)(1-q^{m-1})\cdots(1-q^{m-r+1})} {(1-q)(1-q^2)\cdots(1-q^r)} & r \le m \\ 0 & r>m \end{cases}} </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi></mrow></msub><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mrow><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mtable displaystyle="false" rowspacing=".2em"><mtr><mtd><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><msup><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi></mrow></msup><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><msup><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn></mrow></msup><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><msup><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn></mrow></msup><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo></mrow><mrow><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><msup><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn></mrow></msup><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><msup><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi></mrow></msup><mo stretchy="false"> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo></mrow></mfrac></mrow></mtd><mtd><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi></mtd></mtr><mtr><mtd><mn> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mn></mtd><mtd><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi><mo> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mo><mi> ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </mi></mtd></mtr></mtable></mrow></mrow></mstyle></mrow> </math>${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ ${m \choose r}_{q}={\begin{cases}{\frac {(1-q^{m})(1-q^{m-1})\cdots (1-q^{m-r+1})}{(1-q)(1-q^{2})\cdots (1-q^{r})}}&r\leq m\\0&r>m\end{cases}}$ </img>

où m et r sont des entiers naturels. Pour r = 0 la valeur est 1 puisque le numérateur et le dénominateur sont des produits sur l'ensemble vides . Bien que la formule dans le premier cas semble impliquer une fonction rationnelle, elle désigne en fait un polynôme, car la division est exacte dans Z [ q ] . Notez que la formule peut être appliquée pour r = m + 1 et donne à 0 grâce à un facteur 1 − q⁰ = 0 dans le numérateur, conformément au second cas (pour r encore plus grand le facteur 0 reste présent dans le numérateur, mais ses autres facteurs impliqueraient des puissances négatives de q, donc l’énonciation explicite du second cas est préférable). Tous les facteurs du numérateur et du dénominateur sont divisibles par 1 − q, en tant que quotient un q-nombre définit comme suit :

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo stretchy="false"> <math>[k]_q=\sum_{0\leq i<k}q^i=1+q+q^2+\cdots+q^{k-1}= \begin{cases} \frac{1-q^k}{1-q} & \text{for} & q \neq 1 \\ k & \text{for} & q = 1 \end{cases},} </mo><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><msub><mo stretchy="false"> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi></mrow></msub><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><munder><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mrow class="MJX-TeXAtom-ORD"><mn> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mn><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi></mrow></munder><msup><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi></mrow></msup><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mn> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mn><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><msup><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mn></mrow></msup><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><msup><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mn> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mn></mrow></msup><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mrow class="MJX-TeXAtom-ORD"><mrow><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mtable displaystyle="false" rowspacing=".2em"><mtr><mtd><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mn> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mn><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><msup><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi></mrow></msup></mrow><mrow><mn> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mn><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi></mrow></mfrac></mrow></mtd><mtd><mrow class="MJX-TeXAtom-ORD"><mtext> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mtext></mrow></mtd><mtd><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mn> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mn></mtd></mtr><mtr><mtd><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi></mtd><mtd><mrow class="MJX-TeXAtom-ORD"><mtext> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mtext></mrow></mtd><mtd><mi> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mi><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo><mn> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mn></mtd></mtr></mtable></mrow></mrow><mo> $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </mo></mstyle></mrow> </math>$[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ $[k]_{q}=\sum _{0\leq i<k}q^{i}=1+q+q^{2}+\cdots +q^{k-1}={\begin{cases}{\frac {1-q^{k}}{1-q}}&{\text{for}}&q\neq 1\\k&{\text{for}}&q=1\end{cases}},$ </img>

en utilisant cette notation, on obtient la formule équivalente

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q=\frac{[m]_q[m-1]_q\cdots[m-r+1]_q}{[1]_q[2]_q\cdots[r]_q}\quad(r\leq m),} </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi></mrow></msub><mo> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi></mrow></msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi><mo> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mn> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mn><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi></mrow></msub><mo> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi><mo> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi><mo> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mn> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mn><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi></mrow></msub></mrow><mrow><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mn> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mn><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi></mrow></msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mn> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mn><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi></mrow></msub><mo> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi></mrow></msub></mrow></mfrac></mrow><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi><mo> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mi><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo><mo> ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </mo></mstyle></mrow> </math>${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ ${m \choose r}_{q}={\frac {[m]_{q}[m-1]_{q}\cdots [m-r+1]_{q}}{[1]_{q}[2]_{q}\cdots [r]_{q}}}\quad (r\leq m),$ </img>

ce qui rend évident le fait que remplacer q par 1 dans Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr_q} </mo></mrow><mfrac linethickness="0"><mi> ${\tbinom {m}{r}}_{q}$ </mi><mi> ${\tbinom {m}{r}}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="1.2em" minsize="1.2em"> ${\tbinom {m}{r}}_{q}$ </mo></mrow></mrow></mstyle></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${\tbinom {m}{r}}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${\tbinom {m}{r}}_{q}$ ${\tbinom {m}{r}}_{q}$ </img> donne le coefficient binomial ordinaire Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr.} </mo></mrow><mfrac linethickness="0"><mi> ${\tbinom {m}{r}}.$ </mi><mi> ${\tbinom {m}{r}}.$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="1.2em" minsize="1.2em"> ${\tbinom {m}{r}}.$ </mo></mrow></mrow></mstyle></mrow><mo> ${\tbinom {m}{r}}.$ </mo></mstyle></mrow> </math>${\tbinom {m}{r}}.$ ${\displaystyle }$${\tbinom {m}{r}}.$ ${\tbinom {m}{r}}.$ </img> En termes de q-factorielleÉchec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mo stretchy="false"> <math>[n]_q!=[1]_q[2]_q\cdots[n]_q} </mo><mi> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mi><msub><mo stretchy="false"> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mi></mrow></msub><mo> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mo> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mo stretchy="false"> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mn> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mn><msub><mo stretchy="false"> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mi></mrow></msub><mo stretchy="false"> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mn> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mn><msub><mo stretchy="false"> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mi></mrow></msub><mo> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mo stretchy="false"> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mi> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mi><msub><mo stretchy="false"> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </mi></mrow></msub></mstyle></mrow> </math>$[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ $[n]_{q}!=[1]_{q}[2]_{q}\cdots [n]_{q}$ </img> , la formule peut être énoncée comme

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q=\frac{[m]_q!}{[r]_q!\,[m-r]_q!}\quad(r\leq m),} </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi></mrow></msub><mo> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi></mrow></msub><mo> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo></mrow><mrow><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi></mrow></msub><mo> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi><mo> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi><msub><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi></mrow></msub><mo> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo></mrow></mfrac></mrow><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi><mo> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mi> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mi><mo stretchy="false"> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo><mo> ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </mo></mstyle></mrow> </math>${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ ${m \choose r}_{q}={\frac {[m]_{q}!}{[r]_{q}!\,[m-r]_{q}!}}\quad (r\leq m),$ </img>

une forme compacte (souvent considérée comme une simple définition), qui cache cependant la présence de nombreux facteurs communs au numérateur et au dénominateur. Cette forme rend évidente la symétrie Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr_q=\tbinom m{m-r}_q} </mo></mrow><mfrac linethickness="0"><mi> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mi><mi> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="1.2em" minsize="1.2em"> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mo></mrow></mrow></mstyle></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mi></mrow></msub><mo> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mo></mrow><mfrac linethickness="0"><mi> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mi><mrow><mi> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mi><mo> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mo><mi> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mi></mrow></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="1.2em" minsize="1.2em"> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mo></mrow></mrow></mstyle></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ ${\tbinom {m}{r}}_{q}={\tbinom {m}{m-r}}_{q}$ </img> pour r ≤ m .

Contrairement au coefficient binomial ordinaire, le coefficient binominal de gauss prend des valeurs finies lorsque Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math>m\rightarrow \infty} </mi><mo stretchy="false"> $m\rightarrow \infty$ </mo><mi mathvariant="normal"> $m\rightarrow \infty$ </mi></mstyle></mrow> </math>$m\rightarrow \infty$ $m\rightarrow \infty$ </img> (la limite ayant du sens analytiquement pour | q | < 1):

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{\infty \choose r}_q = \lim_{m\rightarrow \infty} {m \choose r}_q = \frac{1}{[r]_q!\,(1-q)^r}} </mo></mrow><mfrac linethickness="0"><mi mathvariant="normal"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi></mrow></msub><mo> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><munder><mo movablelimits="true"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi><mo stretchy="false"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mi mathvariant="normal"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi></mrow></munder><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo></mrow><mfrac linethickness="0"><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi></mrow></msub><mo> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mn> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mn><mrow><mo stretchy="false"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi><msub><mo stretchy="false"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi></mrow></msub><mo> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mo stretchy="false"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mn> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mn><mo> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi><msup><mo stretchy="false"> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </mi></mrow></msup></mrow></mfrac></mrow></mstyle></mrow> </math>${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ ${\infty \choose r}_{q}=\lim _{m\rightarrow \infty }{m \choose r}_{q}={\frac {1}{[r]_{q}!\,(1-q)^{r}}}$ </img>

Exemples

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{0 \choose 0}_q = {1 \choose 0}_q = 1} </mo></mrow><mfrac linethickness="0"><mn> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mn><mn> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mi></mrow></msub><mo> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mo></mrow><mfrac linethickness="0"><mn> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mn><mn> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mi></mrow></msub><mo> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mo><mn> ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </mn></mstyle></mrow> </math>${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ ${0 \choose 0}_{q}={1 \choose 0}_{q}=1$ </img>

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{1 \choose 1}_q = \frac{1-q}{1-q}=1} </mo></mrow><mfrac linethickness="0"><mn> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mn><mn> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mi></mrow></msub><mo> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mn> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mn><mo> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mo><mi> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mi></mrow><mrow><mn> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mn><mo> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mo><mi> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mi></mrow></mfrac></mrow><mo> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mo><mn> ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </mn></mstyle></mrow> </math>${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ ${1 \choose 1}_{q}={\frac {1-q}{1-q}}=1$ </img>

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{2 \choose 1}_q = \frac{1-q^2}{1-q}=1+q} </mo></mrow><mfrac linethickness="0"><mn> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mn><mn> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mi></mrow></msub><mo> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mn> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mn><mo> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mo><msup><mi> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mn></mrow></msup></mrow><mrow><mn> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mn><mo> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mo><mi> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mi></mrow></mfrac></mrow><mo> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mo><mn> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mn><mo> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mo><mi> ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </mi></mstyle></mrow> </math>${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ ${2 \choose 1}_{q}={\frac {1-q^{2}}{1-q}}=1+q$ </img>

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{3 \choose 1}_q = \frac{1-q^3}{1-q}=1+q+q^2} </mo></mrow><mfrac linethickness="0"><mn> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mn><mn> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mi></mrow></msub><mo> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mn> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mn><mo> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mo><msup><mi> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mn></mrow></msup></mrow><mrow><mn> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mn><mo> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mo><mi> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mi></mrow></mfrac></mrow><mo> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mo><mn> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mn><mo> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mo><mi> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mi><mo> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mo><msup><mi> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </mn></mrow></msup></mstyle></mrow> </math>${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ ${3 \choose 1}_{q}={\frac {1-q^{3}}{1-q}}=1+q+q^{2}$ </img>

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{3 \choose 2}_q = \frac{(1-q^3)(1-q^2)}{(1-q)(1-q^2)}=1+q+q^2} </mo></mrow><mfrac linethickness="0"><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mi></mrow></msub><mo> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mo stretchy="false"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn><mo> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><msup><mi> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn></mrow></msup><mo stretchy="false"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mo stretchy="false"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn><mo> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><msup><mi> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn></mrow></msup><mo stretchy="false"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo></mrow><mrow><mo stretchy="false"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn><mo> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mi> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mi><mo stretchy="false"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mo stretchy="false"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn><mo> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><msup><mi> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn></mrow></msup><mo stretchy="false"> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo></mrow></mfrac></mrow><mo> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn><mo> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><mi> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mi><mo> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mo><msup><mi> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </mn></mrow></msup></mstyle></mrow> </math>${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ ${3 \choose 2}_{q}={\frac {(1-q^{3})(1-q^{2})}{(1-q)(1-q^{2})}}=1+q+q^{2}$ </img>

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{4 \choose 2}_q = \frac{(1-q^4)(1-q^3)}{(1-q)(1-q^2)}=(1+q^2)(1+q+q^2)=1+q+2q^2+q^3+q^4} </mo></mrow><mfrac linethickness="0"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi></mrow></msub><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo></mrow><mrow><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo></mrow></mfrac></mrow><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo stretchy="false"> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn><msup><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup></mstyle></mrow> </math>${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ ${4 \choose 2}_{q}={\frac {(1-q^{4})(1-q^{3})}{(1-q)(1-q^{2})}}=(1+q^{2})(1+q+q^{2})=1+q+2q^{2}+q^{3}+q^{4}$ </img>

Description combinatoire

Au lieu de ces expressions algébriques, on peut également donner une définition combinatoire des coefficients binomiaux de gauss. Le coefficient binomial ordinaire Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr} </mo></mrow><mfrac linethickness="0"><mi> ${\tbinom {m}{r}}$ </mi><mi> ${\tbinom {m}{r}}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="1.2em" minsize="1.2em"> ${\tbinom {m}{r}}$ </mo></mrow></mrow></mstyle></mrow></mstyle></mrow> </math>${\tbinom {m}{r}}$ ${\tbinom {m}{r}}$ </img> compte le nombre de combinaisons. C'est à dire le nombre de façon de choisir choisir r éléments dans un ensemble éléments à m éléments . Si l' on prend les m éléments comme les différentes positions de caractère dans un mot de longueur m chaque r -combinaisons correspond à un mot de longueur m en utilisant un alphabet de deux lettres, disons {0,1}, avec r copies du lettre 1 (indiquant les positions dans la combinaison choisie) et m − r lettres 0 (pour les positions restantes).

les Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{4 \choose 2} = 6} </mo></mrow><mfrac linethickness="0"><mn> ${4 \choose 2}=6$ </mn><mn> ${4 \choose 2}=6$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${4 \choose 2}=6$ </mo></mrow></mrow></mrow><mo> ${4 \choose 2}=6$ </mo><mn> ${4 \choose 2}=6$ </mn></mstyle></mrow> </math>${4 \choose 2}=6$ ${4 \choose 2}=6$ </img> mots utilisant des 0 et des 1 seraient 0011, 0101, 0110, 1001, 1010, 1100.

Pour obtenir de ce modèle le coefficient binomial de Gauss Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="false" scriptlevel="0"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="1.2em" minsize="1.2em"> <math>\tbinom mr_q} </mo></mrow><mfrac linethickness="0"><mi> ${\tbinom {m}{r}}_{q}$ </mi><mi> ${\tbinom {m}{r}}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="1.2em" minsize="1.2em"> ${\tbinom {m}{r}}_{q}$ </mo></mrow></mrow></mstyle></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${\tbinom {m}{r}}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${\tbinom {m}{r}}_{q}$ ${\tbinom {m}{r}}_{q}$ </img> , il suffit de compter chaque mot avec un facteur q^d, où d est le nombre "d'inversions" dans le mot mot: le nombre de paires de positions pour lesquelles la position la plus à gauche de la paire contient la lettre 1 et la position la plus à droite une lettre 0 dans le mot. Par exemple, il y a un mot avec 0 inversions, 0011. Il y a 1 avec une seule inversion, 0101. Il y a deux mots avec 2 inversions, 0110 et 1001. Il y en a un avec 3, 1010, et enfin un mot avec 4 inversions, 1100. Ceci correspond aux coefficients de Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{4 \choose 2}_q = 1+q+2q^2+q^3+q^4} </mo></mrow><mfrac linethickness="0"><mn> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mn> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mi></mrow></msub><mo> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mn><mo> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mi> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mo> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mo><mn> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mn><msup><mi> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup><mo> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mo><msup><mi> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </mn></mrow></msup></mstyle></mrow> </math>${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ ${4 \choose 2}_{q}=1+q+2q^{2}+q^{3}+q^{4}$ </img> .

On peut montrer que les polynômes ainsi définis satisfont les identités de Pascal données ci-dessous et coïncident donc avec les polynômes donnés par les définitions algébriques. Une manière visuelle de comprendre cette définition consiste à associer à chaque mot un chemin traversant une grille rectangulaire avec des côtés de hauteur r et de largeur m − r, du coin inférieur gauche au coin supérieur droit, en faisant un pas à droite pour chaque lettre un pas en avant pour chaque lettre 1. Ensuite, le nombre d'inversions du mot est égal à l'aire de la partie du rectangle située en bas à droite du tracé.

Propriétés

Comme les coefficients binomiaux ordinaires, les coefficients binomaux de Gauss sont symétriques, c'est-à-dire invariants par l'application Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math> r \rightarrow m-r } </mi><mo stretchy="false"> $r\rightarrow m-r$ </mo><mi> $r\rightarrow m-r$ </mi><mo> $r\rightarrow m-r$ </mo><mi> $r\rightarrow m-r$ </mi></mstyle></mrow> </math>$r\rightarrow m-r$ $r\rightarrow m-r$ </img>  :

<mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mi><mi> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mi></mrow></msub><mo> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mi><mrow><mi> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mi><mo> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mo><mi> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mi></mrow></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mi></mrow></msub><mo> ${m \choose r}_{q}={m \choose m-r}_{q}.$ </mo></mstyle></mrow> ${m \choose r}_{q}={m \choose m-r}_{q}.$ ${\displaystyle }$${m \choose r}_{q}={m \choose m-r}_{q}.$ ${m \choose r}_{q}={m \choose m-r}_{q}.$ </img>

En particulier,

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose 0}_q ={m \choose m}_q=1 \, ,} </mo></mrow><mfrac linethickness="0"><mi> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mi><mn> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mi></mrow></msub><mo> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mo></mrow><mfrac linethickness="0"><mi> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mi><mi> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mi></mrow></msub><mo> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mo><mn> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mn><mo> ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </mo></mstyle></mrow> </math>${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ ${m \choose 0}_{q}={m \choose m}_{q}=1\,,$ </img>

<mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo></mrow><mfrac linethickness="0"><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi><mn> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi></mrow></msub><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo></mrow><mfrac linethickness="0"><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi><mrow><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mn> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mn></mrow></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi></mrow></msub><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><mn> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mn><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><msup><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi></mrow></msup></mrow><mrow><mn> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mn><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi></mrow></mfrac></mrow><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mn> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mn><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><msup><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mn> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mn></mrow></msup><mi> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mi><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo><mn> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mn><mo> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </mo></mstyle></mrow> ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ ${\displaystyle }$${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ ${m \choose 1}_{q}={m \choose m-1}_{q}={\frac {1-q^{m}}{1-q}}=1+q+\cdots +q^{m-1}\quad m\geq 1\,.$ </img>

Le nom coefficient binomial de Gauss vient du fait ^{[réf. nécessaire]} que leur évaluation à q = 1 est

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><munder><mo movablelimits="true"> <math>\lim_{q \to 1} {m \choose r}_q = {m \choose r}} </mo><mrow class="MJX-TeXAtom-ORD"><mi> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mi><mo stretchy="false"> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mo><mn> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mn></mrow></munder><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mo></mrow><mfrac linethickness="0"><mi> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mi><mi> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mi></mrow></msub><mo> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mo><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mo></mrow><mfrac linethickness="0"><mi> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mi><mi> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </mo></mrow></mrow></mrow></mstyle></mrow> </math>$\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ $\lim _{q\to 1}{m \choose r}_{q}={m \choose r}$ </img>

pour tous m et r .

Les analogues des identités Pascal pour les coefficients binomiaux de Gauss sont

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q = q^r {m-1 \choose r}_q + {m-1 \choose r-1}_q} </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi></mrow></msub><mo> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo><msup><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi></mrow></msup><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo></mrow><mfrac linethickness="0"><mrow><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi><mo> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo><mn> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mn></mrow><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi></mrow></msub><mo> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo></mrow><mfrac linethickness="0"><mrow><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi><mo> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo><mn> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mn></mrow><mrow><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi><mo> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo><mn> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mn></mrow></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ ${m \choose r}_{q}=q^{r}{m-1 \choose r}_{q}+{m-1 \choose r-1}_{q}$ </img>

et

<mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi></mrow></msub><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo></mrow><mfrac linethickness="0"><mrow><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo><mn> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mn></mrow><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi></mrow></msub><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo><msup><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi></mrow></msup><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo></mrow><mfrac linethickness="0"><mrow><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo><mn> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mn></mrow><mrow><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo><mn> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mn></mrow></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mi></mrow></msub><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </mo></mstyle></mrow> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ ${\displaystyle }$${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}.$ </img>

Il existe des analogues de la formule du binôme de Newton et de la version généralisée de celle-ci pour les exposants entiers négatifs, bien que, dans le premier cas, les coefficients binomiaux de gauss eux-mêmes n'apparaissent pas sous forme de coefficients:

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><munderover><mo> <math>\prod_{k=0}^{n-1} (1+q^kt)=\sum_{k=0}^n q^{k(k-1)/2} {n \choose k}_q t^k } </mo><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mo> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mn> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mn></mrow><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mo> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mn> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mn></mrow></munderover><mo stretchy="false"> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mn> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mn><mo> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><msup><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi></mrow></msup><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mo stretchy="false"> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mo> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><munderover><mo> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mo> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mn> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mn></mrow><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi></mrow></munderover><msup><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mo stretchy="false"> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mo> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mn> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mn><mo stretchy="false"> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo><mrow class="MJX-TeXAtom-ORD"><mo> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo></mrow><mn> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mn></mrow></msup><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo></mrow><mfrac linethickness="0"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi></mrow></msub><msup><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </mi></mrow></msup></mstyle></mrow> </math>$\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ $\prod _{k=0}^{n-1}(1+q^{k}t)=\sum _{k=0}^{n}q^{k(k-1)/2}{n \choose k}_{q}t^{k}$ </img>

et

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><munderover><mo> <math>\prod_{k=0}^{\infty} (1+q^kt)=\sum_{k=0}^\infty \frac{q^{k(k-1)/2}t^k}{[k]_q!\,(1-q)^k} } </mo><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mn> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mn></mrow><mrow class="MJX-TeXAtom-ORD"><mi mathvariant="normal"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi></mrow></munderover><mo stretchy="false"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mn> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mn><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><msup><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi></mrow></msup><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><mo stretchy="false"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><munderover><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mn> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mn></mrow><mrow class="MJX-TeXAtom-ORD"><mi mathvariant="normal"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi></mrow></munderover><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow><msup><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><mo stretchy="false"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mn> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mn><mo stretchy="false"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo></mrow><mn> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mn></mrow></msup><msup><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi></mrow></msup></mrow><mrow><mo stretchy="false"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><msub><mo stretchy="false"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi></mrow></msub><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mo stretchy="false"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mn> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mn><mo> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi><msup><mo stretchy="false"> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mo><mrow class="MJX-TeXAtom-ORD"><mi> $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </mi></mrow></msup></mrow></mfrac></mrow></mstyle></mrow> </math>$\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ $\prod _{k=0}^{\infty }(1+q^{k}t)=\sum _{k=0}^{\infty }{\frac {q^{k(k-1)/2}t^{k}}{[k]_{q}!\,(1-q)^{k}}}$ </img>

qui, lorsque Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math>n\rightarrow\infty} </mi><mo stretchy="false"> $n\rightarrow \infty$ </mo><mi mathvariant="normal"> $n\rightarrow \infty$ </mi></mstyle></mrow> </math>$n\rightarrow \infty$ $n\rightarrow \infty$ </img> devient:

et

$\prod _{k=0}^{\infty }{\frac {1}{(1-q^{k}t)}}=\sum _{k=0}^{\infty }{\frac {t^{k}}{[k]_{q}!\,(1-q)^{k}}}.$

La première identité de Pascal permet de calculer les coefficients binomiaux de gauss de manière récursive (par rapport à m ) en utilisant les valeurs initiales "frontière"

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose m}_q ={m \choose 0}_q=1 } </mo></mrow><mfrac linethickness="0"><mi> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mi><mi> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mi></mrow></msub><mo> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mo></mrow><mfrac linethickness="0"><mi> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mi><mn> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mi></mrow></msub><mo> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mo><mn> ${m \choose m}_{q}={m \choose 0}_{q}=1$ </mn></mstyle></mrow> </math>${m \choose m}_{q}={m \choose 0}_{q}=1$ ${m \choose m}_{q}={m \choose 0}_{q}=1$ </img>

et montre également que les coefficients binomiaux de Gauss sont bien des polynômes (en q ). La seconde identité Pascal découle de la première en faisant le changement Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math> r \rightarrow m-r } </mi><mo stretchy="false"> $r\rightarrow m-r$ </mo><mi> $r\rightarrow m-r$ </mi><mo> $r\rightarrow m-r$ </mo><mi> $r\rightarrow m-r$ </mi></mstyle></mrow> </math>$r\rightarrow m-r$ $r\rightarrow m-r$ </img> et en utilisant l'invariance des coefficients binomiaux de Gauss par le changement de variable Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math> r \rightarrow m-r } </mi><mo stretchy="false"> $r\rightarrow m-r$ </mo><mi> $r\rightarrow m-r$ </mi><mo> $r\rightarrow m-r$ </mo><mi> $r\rightarrow m-r$ </mi></mstyle></mrow> </math>$r\rightarrow m-r$ $r\rightarrow m-r$ </img> . Les deux identités de Pascal impliquent ensemble

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q = {{1-q^{m}}\over {1-q^{m-r}}} {m-1 \choose r}_q } </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi></mrow></msub><mo> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mo><mrow class="MJX-TeXAtom-ORD"><mfrac><mrow class="MJX-TeXAtom-ORD"><mn> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mn><mo> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mo><msup><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi></mrow></msup></mrow><mrow class="MJX-TeXAtom-ORD"><mn> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mn><mo> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mo><msup><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi><mo> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mo><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi></mrow></msup></mrow></mfrac></mrow><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mo></mrow><mfrac linethickness="0"><mrow><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi><mo> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mo><mn> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mn></mrow><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ ${m \choose r}_{q}={{1-q^{m}} \over {1-q^{m-r}}}{m-1 \choose r}_{q}$ </img>

ce qui conduit (lorsqu'il est appliqué de manière itérative pour m, m - 1, m - 2, ...) à une expression du coefficient binomial de Gauss, telle que donnée dans la définition ci-dessus.

Applications

Les coefficients binomiaux de Gauss apparaissent dans le dénombrement des polynômes symétriques et dans la théorie des partitions . Le coefficient en q ^r en

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{n+m \choose m}_q} </mo></mrow><mfrac linethickness="0"><mrow><mi> ${n+m \choose m}_{q}$ </mi><mo> ${n+m \choose m}_{q}$ </mo><mi> ${n+m \choose m}_{q}$ </mi></mrow><mi> ${n+m \choose m}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${n+m \choose m}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${n+m \choose m}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${n+m \choose m}_{q}$ ${n+m \choose m}_{q}$ </img>

est le nombre de partitions de r parmis m ou moins de parties inférieures ou égales à n . De manière équivalente, il s'agit également du nombre de partitions de r parmis n ou moins de parties inférieures ou égales à m .

Les coefficients binomiaux de gauss jouent également un rôle important dans la théorie du dénombrement des espaces projectifs définis sur un corps fini. En particulier, pour chaque corps fini F _q avec q éléments, le coefficient binomial de gauss

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{n \choose k}_q} </mo></mrow><mfrac linethickness="0"><mi> ${n \choose k}_{q}$ </mi><mi> ${n \choose k}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${n \choose k}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${n \choose k}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${n \choose k}_{q}$ ${n \choose k}_{q}$ </img>

compte le nombre de sous-espaces vectoriels de dimension k d'un espace vectoriel de dimension n sur F _q (un grassmannien ). Lorsqu'il est étendu en tant que polynôme en q, il produit la décomposition bien connue du grassmannien en cellules de Schubert. Par exemple, le coefficient binomial de gauss

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{n \choose 1}_q=1+q+q^2+\cdots+q^{n-1}} </mo></mrow><mfrac linethickness="0"><mi> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mi><mn> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mn></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mi></mrow></msub><mo> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mo><mn> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mn><mo> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mo><mi> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mi><mo> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mo><msup><mi> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mn></mrow></msup><mo> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mo><mo> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mo><mo> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mo><msup><mi> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mi><mo> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mo><mn> ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </mn></mrow></msup></mstyle></mrow> </math>${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ ${n \choose 1}_{q}=1+q+q^{2}+\cdots +q^{n-1}$ </img>

est le nombre de sous-espaces de dimension 1 dans ( F _q ) ⁿ (de manière équivalente, le nombre de points dans l' espace projectif associé). De plus, lorsque q vaut 1 (respectivement -1), le coefficient binomial de gauss donne la caractéristique d'Euler du complexe de Grassmann complexe (respectivement réel) correspondant.

Le nombre de sous-espaces affines de dimension k de F _q ⁿ est égal à

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msup><mi> <math>q^{n-k} {n \choose k}_q} </mi><mrow class="MJX-TeXAtom-ORD"><mi> $q^{n-k}{n \choose k}_{q}$ </mi><mo> $q^{n-k}{n \choose k}_{q}$ </mo><mi> $q^{n-k}{n \choose k}_{q}$ </mi></mrow></msup><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> $q^{n-k}{n \choose k}_{q}$ </mo></mrow><mfrac linethickness="0"><mi> $q^{n-k}{n \choose k}_{q}$ </mi><mi> $q^{n-k}{n \choose k}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> $q^{n-k}{n \choose k}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> $q^{n-k}{n \choose k}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>$q^{n-k}{n \choose k}_{q}$ $q^{n-k}{n \choose k}_{q}$ </img> .

Cela permet une autre interprétation de l'identité

Échec de l’analyse (SVG (MathML peut être activé via une extension du navigateur) : réponse non valide(« Math extension cannot connect to Restbase. ») du serveur « http://localhost:6011/fr.wikipedia.org/v1/ » :): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> <math>{m \choose r}_q = {m-1 \choose r}_q + q^{m-r}{m-1 \choose r-1}_q} </mo></mrow><mfrac linethickness="0"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi></mrow></msub><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo></mrow><mfrac linethickness="0"><mrow><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo><mn> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mn></mrow><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi></mrow></msub><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo><msup><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi></mrow></msup><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo></mrow><mfrac linethickness="0"><mrow><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo><mn> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mn></mrow><mrow><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi><mo> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo><mn> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mn></mrow></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><mi> ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </mi></mrow></msub></mstyle></mrow> </math>${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ ${m \choose r}_{q}={m-1 \choose r}_{q}+q^{m-r}{m-1 \choose r-1}_{q}$ </img>

en comptant les sous - espaces de dimension (r - 1) de l'espace projectif de dimension (m - 1), en fixant un hyperplan, en comptant les sous - espaces contenus dans ce hyperplan, puis en comptant les sous - espaces qui ne figurent pas dans l'hyperplan; ces derniers sous-espaces sont en correspondance bijective avec les sous-espaces affines de dimensions ( r - 1) de l'espace obtenu en traitant cet hyperplan fixe comme l'hyperplan à l'infini.

Dans les conventions courantes dans les applications aux groupes quantiques, une définition légèrement différente est utilisée; le coefficient binomial quantique il y a

Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msup><mi> <math>q^{k^2 - n k}{n \choose k}_{q^2}} </mi><mrow class="MJX-TeXAtom-ORD"><msup><mi> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mn></mrow></msup><mo> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mo><mi> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mi><mi> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mi></mrow></msup><msub><mrow class="MJX-TeXAtom-ORD"><mrow><mrow class="MJX-TeXAtom-OPEN"><mo maxsize="2.047em" minsize="2.047em"> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mo></mrow><mfrac linethickness="0"><mi> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mi><mi> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mi></mfrac><mrow class="MJX-TeXAtom-CLOSE"><mo maxsize="2.047em" minsize="2.047em"> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mo></mrow></mrow></mrow><mrow class="MJX-TeXAtom-ORD"><msup><mi> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mi><mrow class="MJX-TeXAtom-ORD"><mn> $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </mn></mrow></msup></mrow></msub></mstyle></mrow> </math>$q^{k^{2}-nk}{n \choose k}_{q^{2}}$ $q^{k^{2}-nk}{n \choose k}_{q^{2}}$ </img> .

Cette version du coefficient binomial quantique est invariante par le changement de Échec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><mi> <math>q} </mi></mstyle></mrow> </math>$q$ $q$ </img> enÉchec de l’analyse (erreur de syntaxe): {\displaystyle <mrow class="MJX-TeXAtom-ORD"><mstyle displaystyle="true" scriptlevel="0"><msup><mi> <math>q^{-1}} </mi><mrow class="MJX-TeXAtom-ORD"><mo> $q^{-1}$ </mo><mn> $q^{-1}$ </mn></mrow></msup></mstyle></mrow> </math>$q^{-1}$ $q^{-1}$ </img> .

Triangles

Les coefficients binomiaux de gauss peuvent être disposés en triangle pour chaque q, pour q=1, on retrouve le triangle de Pascal . </br> Lisez ligne par ligne ces triangles forment les séquences suivantes dans OEIS :

• A022166 pour q = 2
• A022167 pour q = 3
• A022168 pour q = 4
• A022169 pour q = 5
• A022170 pour q = 6
• A022171 pour q = 7
• A022172 pour q = 8
• A022173 pour q = 9
• A022174 pour q = 10