TITLE: Division of Factorials [binomal coefficients are integers]
QUESTION [45 upvotes]: I have a partition of a positive integer $(p)$. How can I prove that the factorial of $p$ can always be divided by the product of the factorials of the parts?
As a quick example $\frac{9!}{(2!3!4!)} = 1260$ (no remainder), where $9=2+3+4$.
I can nearly see it by looking at factors, but I can't see a way to guarantee it.

REPLY [3 votes]: Let's prove using induction, the special case of two numbers, i.e., the statement that if $p, q$ $\in \mathbb{N}$ then $p!q!|(p+q)!$.
(Assume any new variables introduced below to refer to natural numbers.)
First note that since $(p+1)! = (p+1)p!1!$, the statement is true for $q = 1$ and any $p$ (including $p = 1$). In particular, it is true for $p + q = 1 + 1 = 2$. Let us assume that the statement holds for $p, q$ such that $p + q = n$.
Now, for $p, q$ such that $p + q = n + 1$, we can write 
$$(p + q)! = (p + q)(p + q - 1)!$$ 
$$= p [\underbrace{(p - 1) + q}_n]! + q [\underbrace{p + (q - 1)}_n]!$$ 
$$= p \underbrace{k_1 (p-1)!q!}_{\text{using induction assumption}} + q \underbrace{k_2 p!(q-1)!}_{\text{using induction assumption}}$$ 
$$= (k_1 + k_2) p! q!$$. Hence the principle of mathematical induction implies the truth of the statement.
Now it is easy to prove the analogous statement for three numbers, i.e. $p!q!r! | (p + q + r)!$, since (using the statement just proven) $(p + q + r)!$ is divisible by $p! (q+r)!$ and $(q + r)!$ is divisible by $q!r!$.
This can be generalised for any number of parts, by induction.