TITLE: Show that this ratio of factorials is always an integer
QUESTION [37 upvotes]: 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).

REPLY [2 votes]: I am unable to suggest a  $\textbf{combinatorial}$  interpretation of the fact that $~Q(m,n)=\dfrac{(2m)!(2n)!}{m!n!(m+n)!}~$ is an integer (and I do not pretend that what follows is original...) but the result itself is a easy consequence of the two simple propositions :
a) $~\forall x\in \mathbb{R}~~\forall y\in \mathbb{R}~~\lfloor x \rfloor+\lfloor x+y \rfloor+\lfloor y \rfloor~\leq~\lfloor 2x \rfloor+\lfloor 2y \rfloor,~$
b) if, for $~p~$ prime and $~m~$ integer $~\geq 1,~$ we name $~v_p(m)~$ the exponent of $~p~$ in the prime decomposition of $~m,~$ then $~v_p(m!)=\sum\limits_{k\geq 1}\Big\lfloor\dfrac{m}{p^k}\Big\rfloor.~$
Indeed, for all prime $~p~$ and all integers $~m,~n~$ $\geq 1,~$
$v_p(Q(m,n))=\sum\limits_{k\geq 1}\Big(\Big\lfloor\dfrac{2m}{p^k}\Big\rfloor+\Big\lfloor\dfrac{2n}{p^k}\Big\rfloor-\Big\lfloor\dfrac{m}{p^k}\Big\rfloor-\Big\lfloor\dfrac{n}{p^k}\Big\rfloor-\Big\lfloor\dfrac{m+n}{p^k}\Big\rfloor\Big)~\geq 0,$ what was to be shown.