Definition A multinomial coefficient is:
$$ \binom{n}{n_1 n_2 \cdots n_t} =\frac{n !}{n_{1} ! n_{2} ! \cdots n_{t} !} $$
Here, $n_1, n_2, \ldots, n_t$ are nonnegative integers with $$ n_1+n_2+\cdots+n_t=n $$
Pascal’s theorem for multinomial coefficients
Pascal’s theorem for Let $n, k \in \mathbb{N}$ and $(n)_{i=1}^k$ be natural numbers so that $$ n_1+\cdots+n_k=n $$ Then,
$$ \binom{n}{n_1, n_2, \ldots, n_k} = \binom{n-1}{n_1-1, n_2, \ldots, n_k} + \ldots + \binom{n-1}{n_1, n_2, \ldots, n_k-1} $$
Proof
Combinatorial argument: $$ \text { Let } S={\underbrace{1, \ldots, 1}{n_1}, \underbrace{2, \ldots, 2}{n_2}, \ldots, \underbrace{k, \ldots, k}_{n_k}} $$ Count $N = $# of permutations of $S$.
For each $i \in [k]$, define $N_i = $ # of permutations of $S$ where the first element is equal to $i$.
Then, $N = N_1 + N_2 + \ldots + N_k$. Where: $$ N_i = \frac{(n-1)!}{n_1!n_2! \cdots n_{i-1}!(n_i-1)!n_{i+1}\cdots n_k!} $$ Which is equal to the right side. Left side is trivial. Done.
The multinomial theorem
Let $n$ be a positive integer. For all $x_1,\ldots,x_t$, we have
$$ (x_1+\cdots+x_t)^n = \sum \binom{n}{n_1,, n_2, \ldots, , n_k} x_1^{n_1}x_2^{n_2}\cdots x_t^{n_t} $$
where the summation extends over all nonnegative integral solutions $n_1,\ldots,n_t$ of $n_1 + \cdots + n_t = n$.
Proof
Define $S = {n\cdot x_1, n\cdot x_2 \cdots n\cdot x_k}$ ($n$ copies of $x_i$)
Define $\mathcal{P} = $ the collection of all $n$-permutaions of the multiset $S$. $$ (x_1 + \ldots + x_k)^n = \sum_{a_1a_2\ldots a_n \in \mathcal{P}} a_1 \cdot a_2 \cdots a_n $$ Imagine that you just expand this huge multiplication on the left side, then choose one $x_i$ from each parenthesis.
Question: What is the number of $n$-permutation of $S$ with $n_1$ $x_1$’s, $n_2$ $x_2$’s, … $n_k$ $x_k$’s, where $n_1 + n_2 + \ldots + n_k = n$.
Answer: $\binom{n}{n_1, n_2, \ldots, n_k}$
Then we have: $$ (x_1 + \ldots + x_k)^n = \sum_{a_1a_2\ldots a_n \in \mathcal{P}} a_1 \cdot a_2 \cdots a_n = \text{right side of the theorem} $$