# Proof with multinomial.

Let $p$ be a prime number. Prove that $p$ divides the multinomial $$\binom {p}{n_1,n_2,\dots, n_k}$$ such that $n_i \neq p$.

I tried some approaches but honestly i have no idea what to do.

• Can you prove that if $a/b$ is an integer, and $p$ divides $a$ but not $b$, then $p$ divides $a/b$? – Mike Nov 18 '14 at 21:14

This follows directly from the formula for the multinomial coefficient: $$\binom{p}{n_1 ~ n_2 ~ \cdots ~ n_k} = \frac{p!}{n_1!\,n_2!\cdots n_k!}$$ if you also know that it is always an integer.
(The denominator cannot have $p$ as a prime factor).
We know that $$\binom {p}{n_1,n_2,\dots, n_k} = \frac{p!}{n_1!n_2!...n_k!} = \frac{p.(p-1)!}{n_1!n_2!...n_k!}$$
Then it should be clear that $$\frac{\frac{p.(p-1)!}{n_1!n_2!...n_k!}}{p} = \frac{(p-1)!}{n_1!n_2!...n_k!}$$ which is an integer(since $p$ doesn't divide the denominator) and you are done.