Let A $\in$ $M_{n}$ (a matrix) with the numbers 1, 2, ..., $n^2$ in it (each number appears once).
How many matrices B $\in$ $M_{n}$ with each of the numbers 1, 2,..., $n^2$ once there are such that there is no row in B which is the same like some row in A.
I saw similar question here: Inclusion–exclusion: Matrices which also involves columns, but I dont need the columns restriction and the answer given there is not using inclusion-exclusion and not so clear to me.
I take is that when you say that no row of $B$ is the same as (not "same like") some row of $A$ you mean that if the fourth row of $B$ is the same as the first row of $A$, then $B$ is excluded.
This is a straightforward application of inclusion-exclusion; we simply have to count how many $n\times n$ matrices with the elements $1,2,\dots,n^2$ have $k$ rows in common with $A$. There are ${n\choose k}$ ways to choose which rows of $A$ occur in $B$, and ${n\choose k}$ places they can occur in $B$. They can be arranged in $B$ in $k!$ orders. There remain $n^2-kn$ elements in B, and these can be arranged in $(n^2-kn)!$ ways, so altogether we have $${n\choose k}^2k!(n^2-kn)!$$ ways.
I don't think you'll have any trouble completing it from here.