show the formula always gives an integer

$\frac{(2m)!(2n)!}{m!n!(m+n)!}$

I don't remember where I read this problem, but it said this can be proved using a simple counting argument (like observing that $\frac{(3m)!}{m!m!m!}$ is just the number of ways of permuting m identical things of type 1, m of type-2 and m of type-3).