I have to prove that $$\frac{(2n)!(2m)!}{n!m!(n+m)!}$$ is always an integer.
I already have seen the same question here-Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.
But, I am again asking this question as I have a different method from that posted as answers and I am stuck mid way in my proof.So,I want some help to complete this proof. So, please don't close my question as a duplicate.
My attempt-
$$\frac{(2n)!(2m)!}{n!m!(n+m)!}$$
$$=\frac{n!(n+1)(n+2)...(n+n)(m!)(m+1)(m+2)...(m+m)}{n!m!(n+m)!}$$
$$=\frac{(n+1)(n+2)...(n+n)(m+1)(m+2)...(m+m)}{(n+m)!}$$
Case 1:-Let $n>m$
$$=\frac{(n+1)(n+2)...(n+m)(n+m+1)...(n+n)(m+1)(m+2)...(m+m)}{n!(n+1)(n+2)...(n+m)}$$
$$=\frac{(n+m+1)(n+m+2)...(n+n)(m+1)(m+2)...(m+m)}{n!}$$
Now,from $(n+m+1)$ to $(n+n)$, we have $n-m$ terms and from $(m+1)$ to $(m+m)$ we have $m$ terms.So,total we have $(n-m)+m=n$ terms.
Apparently, it seemed to me that this will be an integer as product of any $n$ consecutive integers is always divisible by $n!$.But then to may horror I found out that the $n$ terms in the numerator may not be consecutive.
So, how to get a way out of this by using be method (please, by using my approach, not by any other way!!)?
Thanks a lot in advance!!
It is enough to show that $\binom{n+m}{n}=\binom{n+m}{m}$ is a divisor of $\binom{2n}{n}\cdot\binom{2m}{m}$.
Assume that we have a parliament with $2n$ politicians in the left wing and $2m$ politicians in the right wing. $\binom{2n}{n}\binom{2m}{m}$ is the number of committees made by exactly $n$ politicians from the left wing and exactly $m$ politicians from the right wing. Every subset of $n$ politicians in the left wing has the same probability to be part of the super-committee, and $$ \binom{n+m}{n}\mid \binom{2n}{n}\binom{2m}{m} $$ also comes from computing the probability for a single politician to be part of the super-committee.