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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.

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

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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.

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