Multiplicative Convolution for Binomial Coefficients I know Vandermonde's convolution for binomial coefficients:
$$\sum_{j=0...k} \binom{n}{j} \binom{m}{k-j} = \binom{n+m}{k}$$
Is there a similar multiplicative convolution?  More precisely, is there a simple formula for the coefficients $a(k_j,k'_j,k)$ in the following
identity?
$$\sum_{j} a(k_j,k'_j,k) \binom{n}{k_j} \binom{m}{k'_j} = \binom{n \cdot m}{k}$$ 
Thanks.
 A: The interpretation darij gives in the comments has a fairly straightforward proof: ${nm \choose k}$ is precisely the number of $n \times m$ matrices of $0$s and $1$s with exactly $k$ $1$s. One can think of the terms ${n \choose k_j}$ and ${m \choose k_j'}$ as picking out the rows and columns that these $1$s inhabit, and then the coefficient $a(k_j, k_j', k)$ describes precisely the number of ways to distribute the $1$s given these constraints.
(Note that the coefficients $a(k_j, k_j', k)$ are unique, for example by repeatedly applying finite difference operators in $m$ and $n$.)
A: By a simple Inclusion-Exclusion argument, the number of $r\times s$ (0,1)-matrices with $k$ 1's and with at least one 1 in every row and column is the coefficient of $x^k$ in
  $$ \sum_{j=0}^s (-1)^j{s\choose j}((1+x)^{s-j}-1)^r. $$
This gives a formula for $a(k_j,k'_j,k)$ as a triple sum.
A: Riordan and Stein, in "Arrangements on Chessboards" (Journal of Combinatorial Theory, Series A, 12 72-80, 1972) consider the numbers $A(r,s,k)$ defined by 
$$\sum_{r,s} \binom{n}{r} \binom{m}{s} A(r,s,k) = \binom{nm}{k},$$ or, as others have pointed out, the number of $r \times s$ $(0,1)$-matrices with $k$ $1$'s and with at least one $1$ in every row and every column.  They obtain two formulas for these numbers:
$$A(r,s,k) = \sum_{i,j} (-1)^{i+j+r+s} \binom{r}{i} \binom{s}{j} \binom{ij}{k},$$
$$A(r,s,k) = \sum_n \frac{r! s!}{k!} S(n,r) S(n,s) s(k,n),$$
where $S(n,r)$ and $s(k,n)$ are Stirling numbers of the second and first kinds, respectively.
They also obtain the recurrence relation
$$(k+1)A(r,s,k+1) +k A(r,s,k) $$
$$= rs \left(A(r,s,k) + A(r-1,s,k) + A(r,s-1,k) + A(r-1,s-1,k)\right)$$ 
and that $A(r,s,k)$ is the coefficient of $t^k$ in
$$\sum_j (-1)^{s+j} \binom{s}{j} \left( (1+t)^j-1) \right)^r,$$
as in Richard Stanley's answer.
