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.



Multiplicative Convolution for Binomial Coefficients
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.


