Number of $\{0,1\}$ matrices with distinct rows and distinct columns How many $M\in\{0,1\}^{r\times c}$ are there such that each row and each column of $M$ is distinct?
How many classes of matrices in $\{0,1\}^{r\times c}$ up to permutation equivalence are there such that each matrix in every class has each row and each column distinct?
What is the analogous count if rank over $\Bbb K$ is restricted to exactly $m$ where $\Bbb K$ is a field?
If $\Bbb K=\Bbb R$ then if $r=c$ then asymptotically I think we should have $2^{cm}$ matrices.
 A: This is OEIS sequence A181230. The square case $r=c$ is OEIS sequence A088310.  See those pages for formulas.  As Pat Devlin mentions, the asymptotic problem is trivial if both $r$ and $c$ increase quickly enough. 
A: Ok.  Let's say $r \leq c$ and say $N = 2^r$.
Then the number of matrices with each column distinct is exactly $N(N-1)\cdots (N-c+1)$.  This is an upper bound on the first question you asked.
Approximation time:  suppose $r^2 \ll 2^c$ then almost all (asymptotically all) matrices will have distinct rows (birthday problem) anyway.  Since $r \leq c$, this condition will happen provided $r \gg 1$.  Therefore, having distinct rows basically always happens, so the upper bound given above is essentially the truth [this can be made more rigorous and exact as desired].
If you want to look at the equivalence class thing, then just use the fact that most matrices will have only one automorphism anyway.  So just divide the answer by $r! c!$.
Punchline: the thing is asymptotically easy, and the answer is more or less exactly what you'd think.
