How to prove the following identity?


Let $r = (r\_1, r\_2, \ldots, r\_d)$ and $c = (c\_1, c\_2, \ldots, c\_d)$ be sequences of natural numbers such that $s = r\_1 + r\_2 + \cdots + r\_d = c\_1 + c\_2 + \ldots + c\_d$.


Denote by $\mathcal{M}(r,c)$ the set of matrices whose rows sums and column sums are $r$ and $c$ respectively.


Then
$$
\sum\_{M = (m\_{i,j}) \in \mathcal{M}(r,c)} \prod\_{i,j} \frac{1}{m\_{i,j}!} = \frac{s!}{(\prod\_{i=1}^d r\_i!) (\prod\_{j=1}^d c\_j!)}.
$$


I remember that once I saw this identity, but I could not find the source now.



Combinatorial identity concerning integral matrices with prescribed row sums and column sums
Here is a quick way to see this with generating functions. By using the notation $[\mathbb x^r]f(\mathbb x)$ for the coefficient of a monomial $\mathbb x^r$ in a formal power series $f(\mathbb x)$, we can write:


$$LHS=\left[\prod\_{i,j}x\_i^{r\_i}y\_j^{c\_j}\right]\prod\_{i,j}e^{x\_iy\_j}=\left[\prod\_{i,j}x\_i^{r\_i}y\_j^{c\_j}\right]e^{(x\_1+\cdots+x\_d)(y\_1+\cdots+y\_d)}$$
$$=\frac{1}{s!}\binom{s}{r\_1,r\_2,\dots,r\_d}\binom{s}{c\_1,c\_2,\dots,c\_d}=RHS$$


