TITLE: On the impossibility of proving certain problems using double counting.
QUESTION [11 upvotes]: 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

REPLY [6 votes]: 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.