TITLE: Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.
QUESTION [14 upvotes]: Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.

I'm not familiar to factorial and I don't have much idea, can someone show me how to prove this? Thank you.

REPLY [9 votes]: Denote $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ by $S(n,m)$. It satisfies relation $S(n+1,m)+S(n,m+1)=4S(n,m)$ (cf. Pascal's triangle) — or, equivalently, $S(n+1,m)=4S(n,m)-S(n,m+1)$. Since $S(0,m)=\binom{2m}m$ is an integer, this relation implies that all $S(n,m)$ are integers.
Ref.: I. Gessel. Super Ballot Numbers (via MO)