TITLE: An example where $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is the number of ways of counting something?
QUESTION [24 upvotes]: Prove that for all non-negative integers $m,n$, $\frac{(2m)!(2n)!}{m!n!(m + n)!}$ is an integer.
There is a answer given here to this question here.
I've seen how it can be proven using recurrence equations as well. When I was trying to solve this problem I was looking to represent it as the number of ways of counting something. Can anyone give such an example, thus proving the question another and very natural way?

REPLY [3 votes]: Following from Grigory M's advice here is a way of interpreting these super catalan numbers as counting lattice paths. We also use part of a proof by Vladimir Nikolin when we deal with the Vandermonde Convolution Formula and Szily's identity.
Szily's identity is proven by equating the $x^{2m} $coefficient in the following expression $(1+x)^{m+n}(1-x)^{m+n}=(1-x^2)^{m+n}$ and then multiplying both sides by $\frac{(-1)^m(2m)!(2n)!}{(m+n)!^2}$ 
We get
$$
S(m,n)=\sum_k(-1)^k\binom{2m}{m+k}\binom{2n}{n-k}
$$
This shows that this expression we are dealing with is an integer. It even seems to me like this is close enough to counting something, the number of ways of picking 1s from bracket expansions multiplied by a constant or something. But we can do better.
Grigory M suggested considering the Vandermonde convolution formula;
$$
\sum_k\binom{2m}{m+k}\binom{2n}{n-k}=\binom{2(n+m)}{n+m}.
$$
where the sum is over all k.
This identity can be rewritten as 
$$
\sum_{k=0}\binom{2m}{k}\binom{2n}{m+n-k}
$$
by letting m+k go to k
Our super catalan expression is transformed into $$
S(m,n)=\sum_{k=0}(-1)^{k-m}\binom{2m}{k}\binom{2n}{m+n-k}
$$
Now consider the following proof of the Vandermonde convolution formula;
The number of paths in a lattice from the origin to a point A(m+n, m+n) taking only up and right steps is $\binom{2(m+n)}{m+n}$ (choosing which m+n steps are up steps in the 2(m+n) steps getting there)
Consider the following approach to counting the number of ways;
Look at the line x + y = 2m 
(the diagram is depicting the case of m instead of 2m)

(source: cut-the-knot.org) 
Each path crosses the line at a unique point $A_j$
There are $\binom{2m}{k}$ ways to get to $A_j$
Then there are $\binom{2n}{m+n-k}$ moves left to choose to get to A(m+n, m+n).
So there are $\binom{2m}{k}$$\binom{2n}{m+n-k}$ paths that cross through $A_j$
Summing over all j gives our Vandermonde identity.
Now if we look at our most recent expression for these super catalan numbers we see that it is just counting the number of paths from the origin to A(m+n, m+n) that go through an even $A_i$ minus the number that go through an odd $A_i$ or vice versa depending on the parity of m. We have interpreted this expression as counting something. It follows that it's an integer.
Although I can no longer see at all why it has to be a positive integer, but sure hey, that's what the original expression is for.