Prove that
$$ \dfrac{(2m)! \cdot (2n)!}{(m)! \cdot (n)! \cdot (m+n)!} $$
is a positive integer, where $(m,n) \in \mathbb{Z^{+}}$
I have already solved it using Legendre's Formula which states that $$e_{p}(n)=\sum_{i=1}^{\infty} \bigg\lfloor \dfrac{n}{p^{i}} \bigg\rfloor$$
where $e_{p}(n)$ is the exponent of a prime $p$ in $n!$. For the problem it was sufficient to show that
$$ e_{p}(2m) + e_{p}(2n) \ge e_{p}(m) + e_{p}(n) + e_{p}(m+n) $$
which I can show using the properties of floor function.
However, I'm seeking a combinatorial approach to this problem. For example, using basic combinatorics, I can show that the number of ways to divide $A$ objects into $k$ persons such that the $i^{th}$ person receives $a_{i}$ objects is
$$ \dfrac{A!}{\displaystyle\prod_{i=1}^{k}{(a_{i})!}} = \dfrac{\left(\displaystyle\sum_{i=1}^k (a_{i})\right)!}{\displaystyle\prod_{i=1}^{k}{(a_{i})!}} $$
here, the set $\{a_{i}\}_{i=1}^k$ is exhaustive, i.e,
$ A = \displaystyle\sum_{i=1}^k a_{i} $.
Using this, I can show the following numbers to be integer
$ \dfrac{(2m)! \cdot (2n)!}{[(m)!]^{2} \cdot [(n)!]^{2} } $
$ \dfrac{(2m)! \cdot (2n)!}{(m-n)! \cdot [(n)!]^2 \cdot (m+n)!} $ ; if $m \geq n$
$ \dfrac{(2m)! \cdot (2n)!}{(n-m)! \cdot [(m)!]^2 \cdot (m+n)!} $ ; if $n \geq m$
However, I can't seem to find a way to tackle this problem using my approach.
Edit: I'm specifically asking for an answer using my combinatorics approach as I've already solved it using the answer given in the other question.
Any help will be appreciated.
Thanks.
We just have to prove that $\binom{m+n}{m}$ divides $\binom{2m}{m}\cdot\binom{2n}{n}$.
So, we may imagine to have a parliament, with $2m$ members in the right wing and $2n$ members in the left wing. We may choose a committee with $n$ people from the left wing and $m$ people from the right wing in $\binom{2n}{n}\cdot\binom{2m}{m}$ ways, then we may choose $m$ chiefs of the committee in $\binom{m+n}{n}=\binom{m+n}{m}$ ways.
Now ask yourself: if all the choices are random, what is the probability that a left-wing or a right-wing member of the parliament will be elected chief of the committee?
Can you deduce that $\binom{m+n}{n}$ has to be a divisor of $\binom{2n}{n}\cdot\binom{2m}{m}$?