Suppose $\Phi(m,n)=(2m)!^n\prod\_{k=1}^n\binom{2m+2k+x}{2k+x}$. 
Then, algebraically, it is trivial to see that
$$(2m)!^n\prod\_{k=1}^n\binom{2m+2k+x}{2k+x}=(2n)!^m\prod\_{k=1}^m\binom{2n+2k+x}{2k+x}.$$



> 
> **Question.** Is there a combinatorial (or anything but algebraic) reason why $\Phi(m,n)=\Phi(n,m)$?
> 
> 
> 


**UPDATE.** I still await for any suggestion.



Seeking for a meaning: a curious symmetry
The question is a bit strange, because it would also be true if one removes all the $2$ from the statement. Namely:
$$\Phi'(m, n):=m!^n \prod\_{k = 1} ^ n \binom{m+k+x}{m}$$
is also symmetric.


Let me explain the symmetry of this modified function $\Phi'$, and you can certainly adapt it to your $\Phi$. But the explanation is somewhat artificial.


Assume harmlessly that $x$ is a non-negative integer. Consider a grid of size $m \times n$. The rows and columns are indexed $0, \cdots, m - 1$ and $0, \cdots, n - 1$, respectively, and a cell is indexed $(i, j)$ if it is in row $i$ and column $j$.


Now one wants to fill in the grid with integers, one in a cell, with the following restriction: every integer is at most $m + n + x$, and the integer in the cell $(i, j)$ should be strictly larger than $i + j$.


Here is the claim:


For every $k \in \{0, \cdots, n - 1\}$, the number of different ways to fill in the column $k$ is equal to $$m!\binom{m + n - k + x}{m}.$$
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This explains everything.


To prove the claim, it is (again) algebraically obvious, but if you really want a combinatorial proof, then there is the following bijection:


$$\{(a\_1, \cdots, a\_m): a\_i \in \{1, \cdots, M\}, a\_i \neq a\_j \} \longleftrightarrow \{(c\_1, \cdots, c\_m): c\_i \in \{1, \cdots, M - i\}\},$$


which is described as follows: given such $(a\_i)\_i$, define $c\_i$ to be the integer such that $a\_i$ is the $c\_i$-th largest number in the set $\{1, \cdots, M\} \backslash \{a\_1, \cdots, a\_{i - 1}\}$. The procedure is obviously reversible, hence gives a bijection.


