91Ó°ÊÓ

The Multinomial Theorem is

${\left(x_1+x_2+x_3+......+x_r\right)}^n=\underset{(n_1,n_2,....,n_r):n_1+n_2+...+n_r=n}{∑}\left(\begin{array}{c}n\\ n_1,n_2,n_r\end{array}\right){x_1}^{n_1},{x_2}^{n_2},...,{x_r}^{n_r}$

We have to prove that

${\left(x_1+x_2+x_3+......+x_r\right)}^n=\underset{(n_1,n_2,....,n_r):n_1+n_2+...+n_r=n}{∑}\left(\begin{array}{c}n\\ n_1,n_2,n_r\end{array}\right){x_1}^{n_1},{x_2}^{n_2},...,{x_r}^{n_r}$

Let us expand the L.H.S

${\left(x_1+x_2+x_3+......+x_r\right)}^n=\left(x_1+x_2+x_3+......+x_r\right)×\left(x_1+x_2+x_3+......+x_r\right)×...ntimes$

There will be many terms formed in the final expansion. Terms are formed by

selecting one variable in each$\left(x_1+x_2+x_3+......+x_r\right)$

Thus out of $n$ products of$\left(x_1+x_2+x_3+......+x_r\right)$

$$\begin{gathered} x_{1}canbeselectedfor{n}_{1}times \\ {x}_{2}canbeselectedfor{n}_{2}times \\ . \\ . \\ . \\ {x}_{r}canbeselectedfor{n}_{r}times \end{gathered}$$

Such that$n=n_1+n_2+....+n_r$

To select these, it is equivalent to separating $n$into groups of $n_1,n_2,....,n_r$.

This distribution can be done in$\frac{n!}{n_1!n_2!......n_r!}\text{ways}$

But this is just one term such that

$x_1$appears $n_1$times

$x_2$appears $n_2$times

and so on.

So, the final term is$\left(\begin{array}{c}n\\ n_1,n_2,n_r\end{array}\right){x_1}^{n_1},{x_2}^{n_2},...,{x_r}^{n_r}$

The other terms would be having different $n_1,n_2,....,n_r$and hence the product is a summation of all such terms like (1), where $n=n_1+n_2+....+n_r$

Therefore,${\left(x_1+x_2+x_3+......+x_r\right)}^n=\underset{(n_1,n_2,....,n_r):n_1+n_2+...+n_r=n}{∑}\left(\begin{array}{c}n\\ n_1,n_2,n_r\end{array}\right){x_1}^{n_1},{x_2}^{n_2},...,{x_r}^{n_r}$