Multinomial Theorem

The Multinomial Theorem states that $(a_1+a_2+\cdots+a_k)^n=\sum_{\substack{j_1,j_2,\ldots,j_k \\ 0 \leq j_i \leq n \textrm{ for each } i \\ \textrm{and } j_1 + \ldots + j_k = n}}\binom{n}{j_1; j_2; \ldots ; j_k}a_1^{j_1}a_2^{j_2}\cdots a_k^{j_k}$ where $\binom{n}{j_1; j_2; \ldots ; j_k}$ is the multinomial coefficient $\binom{n}{j_1; j_2; \ldots ; j_k}=\dfrac{n!}{j_1!\cdot j_2!\cdots j_k!}$ .

Note that this is a direct generalization of the Binomial Theorem: when $k = 2$ it simplifies to $(a_1 + a_2)^n = \sum_{\substack{0\leq j_1, j_2 \leq n \\ j_1 + j_2 = n}} \binom{n}{j_1; j_2} a_1^{j_1}a_2^{j_2} = \sum_{j = 0}^n \binom{n}{j} a_1^j a_2^{n - j}$

Proof

Using induction and the Binomial Theorem

We have an arbitrary number of variables to the power of $k$ . For the sake of simplicity, I will use a small example. The problem could be asking for the number of terms in $(x+y+z)^k$ . Since all equivalent parts can be combined, we only need to worry about different variables. Those variables must be in terms of $x$ , $y$ , and $z$ . It's easy to see why the exponent value of each variable must sum to $k$ (Imagine k groups. Pick one variable from each group). Our problem then becomes $a$ + $b$ + $c$ = $k$ , where $a$ , $b$ , and $c$ are the exponents of $x$ , $y$ , and $z$ . This is a straightforward application of the Binomial Theorem. Thus, our answer is ${k+2}\choose{k}$ . A more generalized form would be ${k+(number of variables)-1}\choose{k}$ .