TITLE: Does $a!b!$ always divide $(a+b)!$
QUESTION [10 upvotes]: Hello the question is as stated above and is given to us in the context of group theory, specifically under the heading of isomorphism and products. I would write down what I have tried so far but I have made very little progress in trying to solve this over the last few hours!

REPLY [4 votes]: A proof based on prime factorization:
Let $e(N,p)$, $p$ a prime, be the exponent of $p$ in the prime factorization of $N$. Then we have to show:
For all primes $p$: $e(a!,p)+e(b!,p) \le e((a+b)!,p)$.
We have
$e(n!,p) = [\frac{n}{p}] + [\frac{n}{p^2}] + \cdots, \quad$ where $[r]$ is the largest integer $\le r$.
We also have
$[\frac{a}{k}] + [\frac{b}{k}] \le [\frac{a+b}{k}] \quad$  for any positive integers $a,b,k$.
From these two propositions the claim follows directly.