Factorial of a sum of n numbers divisible by product of factorial of these numbers [duplicate]

Let $a_1, a_2, a_3,...,a_n$ be $n$ positive integers and if $a=a_1+a_2+a_3+...+a_n$. Prove that $a_1!a_2!a_3!...a_n! | a!$

Factorial of a sum of $n$ numbers is divisible by product of factorial of these numbers.

marked as duplicate by Community♦Oct 3 '17 at 16:16

• Hint: how many ways can we permute a collection of $a_1$ apples, $a_2$ oranges, and so on. – lulu Oct 3 '17 at 15:57
$$(x_1 +x_2+\cdots+x_m)^n = \sum \limits_{k_1 +k_2 +\cdots +k_m =n} \binom{n}{k_1,k_2,\cdots,k_m} \prod \limits_{t=1}^{m} x_t^{k_t}$$,
Its easy to see that $$\binom{n}{k_1,k_2,\cdots,k_m}$$ is Integer.