count of matrix of 0,1 in which each row and each column have at least one 1. I know that the number of matrices with $0,1$ entries, in which each row and each column have at least one $1$, is equal to
$$\sum_{r=0}^n \sum_{s=0}^n (-1)^{r+s} C(n,r) C(n,s) 2^{rs}.$$
I want to prove it combinatorially using Inclusion–exclusion principle.

Let's say $S_{t,u}$ is the number of binary $n$ by $n$ matrices with $t$ rows of zeroes and $u$ columns of zeroes. The rows can be chosen in $C(n,t)$ ways and the columns can be chosen in $C(n,u)$ ways. The remaining portion of the matrix has $(n-t)(n-u)$ binary elements. So
$$S_{t,u} = C(n,t)C(n,u)2^{(n-t)(n-u)}$$
(Note that this formula works even when $t=u=0$.)
By inclusion/exclusion, the number of matrices with no row of zeroes and no column of zeroes is
$$\begin{align}
N_0 &= \sum_{t=0}^n \sum_{u=0}^n (-1)^{t+u} S_{t,u} \\
 &= \sum_{t=0}^n \sum_{u=0}^n (-1)^{t+u} C(n,t)C(n,u)2^{(n-t)(n-u)}
\end{align}$$
Now make a change of indices to $r=n-t$ and $s=n-u$.  The result is
$$\begin{align}
N_0 &= \sum_{r=0}^n \sum_{s=0}^n (-1)^{2n-r-s} C(n,n-r)C(n,n-s) 2^{rs} \\
&= \sum_{r=0}^n \sum_{s=0}^n (-1)^{r+s} C(n,r)C(n,s) 2^{rs}
\end{align}$$
