On the impossibility of proving certain problems using double counting.

Usually in combinatorics, I love proofs by double counting. It gives me a very happy feeling to know a double counting proof. I feel I understand the problem better. A close younger sibling of this technique is to interpret a given expression as a solution to a smartly constructed counting problem.

So whenever somebody asks me to prove that a ratio involving factorials is an integer, I try to interpret the ratio as a solution to a counting problem. But throughout my counting life, I have encountered certain expressions which never admit an interpretative proof. One of them is jasoncube's question posted here.

I searched online and I could not find a slick proof for jasoncube's question. So this post has the following two questions:

1) For $m,n \in \mathbb{N}$ can you find a counting problem whose solution is $\dfrac{(2m)! (2n)!}{m! n! (m+n)!}$?

2) Is there any literature on the extent of this technique or it's limitations? That is, has anybody proved impossibility results for certain expressions?

Thanks,
Iso

According to this question on Math Overflow ("Recursions which define polynomials") there is no known combinatorial interpretation of the numbers $$A(m,n) = \frac{(2m)! (2n)!}{m! n! (m+n)!}.$$

The question does link to Ira Gessel's paper "Super Ballot Numbers" (Journal of Symbolic Computation 14 (1992) 179--194). In Section 6 Gessel calls these "super Catalan numbers" and gives a few proofs of their integrality. Equation (32) consists of the formula $$\sum_n 2^{p-2n} \binom{p}{2n} A(m,n) = A(m,m+p), \:\:\: p \geq 0.$$ Gessel says that this formula, together with $A(0,0) = 1$ and $A(m,n) = A(n,m)$, "in principle... gives a combinatorial interpretation to $A(m,n)$, although it remains to be seen whether [the formula] can be interpreted in a 'natural' way."

So "no known combinatorial interpretation, but a recursive formula that might lead to one" appears to be the state of things at this point.