Question

State and prove a multinomial theorem.

Answer 1

For non negative integers $b_1,b_2,.....b_k$ such that $\sum _{1=1}^k{b}_i=n$, the multinomial coefficient is $\left(\begin{array}{c}n\\ b_1,b_2,b_3,.....b_k\end{array}\right)=\frac{n!}{b_1!b_2!b_3!.....b_k!}$

For $k=2$

$\left(\begin{array}{c}n\\ b_1,b_2\end{array}\right)=\frac{n!}{b_1!b_2!}=\frac{n!}{b_1!(n-b_1)!}=\left(\begin{array}{c}n\\ b_1\end{array}\right)$

Theorem:

For positive integers k and non negative integer n,

$(x_1+x_2+...\dots +x_k{)}^n=\sum _{b_1+b_2+b_3+.....+b_k=n}\left(\begin{array}{c}n\\ b_1{b}_2{b}_3.....b_k\end{array}\right)\prod _{j=1}^k{x}_j^{bj}$

The number of terms of this sum are given by $\left(\begin{array}{c}n+k-1\\ n\end{array}\right)$

Theorem:

When $k=1$ result is true, when $k=2$ result in binomial theorem, Assume $k\ge 3$ and the result is true for $k=p$. when $k=P+1$

$(x_1+x_2+x_3.....+x_p+x_{p+1}{)}^n=(x_1+x_2+...\dots +x_{p-1}+(x_p+x_{p+1}){)}^n$

$\sum _{b_1+b_2+.....+b_{p-1}+B=n}\left(\begin{array}{c}n\\ b_1,b_2,.....b_{p-1},B\end{array}\right)\prod _{j=1}^{p-1}{x}_j^{bj}\times (x_p+x_{p+1}{)}^B$

By binomial theorem,

$\sum _{b+1+b_2+....+b_{p+1}=n}\left(\begin{array}{c}n\\ b_1,b_2,.....b_{p-1},B\end{array}\right)\prod _{0=1}^{p-1}{x}_j^{bj}\times \sum _{b_p+b_{p+1}=B}\left(\begin{array}{c}B\\ bp\end{array}\right)x_p^{b_p}{x}_{p+1}^{b_{p+1}}$

$\sum _{b_1+b_2+.....+b_{p+1}=n}\left(\begin{array}{c}n\\ b_1,b_2,....b_{p+1}\end{array}\right)\prod _{0=1}^k{x}_j^{bj}$.