I don't find anything better than just using PIE. Suppose $f(m,n)$ is the number of ways to put rooks on an $m \times n$ board such that every row/column has at least one rook. Then $f(1,n) = f(m,1) = 1$ for all $m,n$, and $f(m,n) = (2^{mn} - 1) - \sum_{i,j} \binom{m}{i} \binom{n}{j} f(i,j)$ where the sum runs over all $1 \le i \le m, 1 \le j \le n$ except $(i,j) = (m,n)$. After this, find $f(8,8)$. This should be easy enough to compute with a program or laboriously by hand, but I'm not sure if there's anything better.

But this is a TST problem... So there ought to be something better. The problem doesn't seem to be good else

Easy enough through complementary counting. Say first column has no rook then there are $2^{56}$ cases. If the second column doesn't have a rook there are $(2^8-1)(2^{48})$ cases that we haven't counted. If the third column doesn't have a rook then there are $(2^8-1)^2(2^{40})$ cases that we haven't counted. etc etc

And what about the condition on the rows? Applying your method for a $2\times 2$ board, we get: "Say first column has no rook, then there are $2^{2}$ cases. If the second column doesn't have a rook, there are $(2^2-1)(2^{0})$ cases that we haven't counted", and it ends here, as there is no third column. Thus there are $2^2 + (2^2-1) = 7$ configurations where at least one column is empty, so there are $2^4-7=9$ configurations where no column is empty. How do you propose to reduce this to the $7$ configurations where also no row is empty? (meaning, in general).

the answer is $2^{64}$ because 0 rooks we can place on 64C0 ways...x rooks we an put on 64Cx ways which is in total 64C0 64C1 +...+64C64 which is by newton formula equal to $2^{64}$ (because $(1+1)^{64}=64C0+...+64C64=2^{64}$ xCy stands for binomial coefficient

Consider this idea. A rook fires along exactly one row and one column. So we have 64 ways to place the first rook. Since there are 8 + 8 -1 = 15 places which cannot be occupied, there are 49 possible squares for the next rook. Moreover, it is easy to see that there are exactly two squares where both the rooks exercise control . So there are 15 + 15 - 2 = 28 squares which cannot be used for the third rook. So there are 36 possible squares for the third rook. Now, there are exactly two squares each where the first and second, second and third, and third and first rook exercise control.Also, there is no square where all three rooks can exercise control, as this would imply that two of them lie on the same row or column, a contradiction. So there are 64 - (15 + 15 + 15 - 2 - 2 - 2) = 25 squares where we can place the fourth rook. Similarly following, we see that there are, finally, 64 .49.36.25.16.9 / 6! =7.8.8! =2257920 ways to cover the board with 8 rooks. Similarly we should consider 7 rooks, 6 rooks, ...., 0 rooks to get the answer.Please correct me if I'm wrong ( My answer seems too large to me, but I'm unable to find any flaws in my proof).

because you counted 1 configuration many times.If you have k rooks you counted it $k!$ times so you actualy have to divide it with $k!$.

also there is no relation of attacking between them

strujabog: You must fill each row/column with at least one rook. You counted the configuration of no rooks, which clearly violates the condition above. AdithyaBhaskar: Same thing; you need to fill each row/column with at least one rook, not at most one rook.

then the problem becomes even more tiresome.

Of course it is ! Else it would already have answers by now

sry i didnt read that

Let $P(m,n)$ be the no of rooks satisfying the condition for a board with $m$ rows and $n$ columns.Then we have the recurrance rel: \[P(m,n)=(2^m-1)^n-(\dbinom{n}{n-1}P(m,n-1)+\dbinom{n}{n-2}P(m,n-2)....\dbinom{n}{1}P(m,1))\] Proof:we consider $S(m,n)$ be the no of ways of arranging rooks such that at least one rook is present in each of $m$ rows.So $S(m,n)=(2^m-1)^n$ From $S(m,n)$ we subtract the ones that have those ways that have one or more columns unoccupied.We would get those ways in which no column is left unoccupied .The no of ways in which $k$ rows are unoccupied is $\binom{n}{n-k}P(m,n-k)$ So with some enumerations we get the result. The solution to the relation i have conjectured to be: \[P(n,n)=(2^n-1)^n-\binom{n}{n-1}(2^{n-1}-1)^{n}+\binom{n}{n-2}(2^{n-2}-1)^{n}......+(-1)^k\binom{n}{n-k}(2^{n-k}-1)^{n}.......+(-1)^{n-1}\binom{n}{1}\] Could someone please check my proof?

@chaotic_iak there should be an easy recursion to compute $\sum_{i+j=k}f(i,j)$

so what is the answer finally ? Anyone ?

First, prove the following version of inclusion-exclusion: If $a(m, n) = \sum_{j = 0}^m \sum_{k = 0}^n \binom{m}{j} \binom{n}{k} b(m, n)$ then $b(m, n) = \sum_{j = 0}^m \sum_{k = 0}^n (-1)^{m+n+j + k} \binom{m}{j} \binom{n}{k} a(m, n)$. Second, apply to this problem in the case $a(m, n) = 2^{mn}$ and $b(m, n) = f(m, n)$. (This was suggested by chaotic_iak back in the very first response.) This double sum can be reduced to a single sum by the binomial theorem, but I doubt it has a simpler closed form. Edit: actually we can also get to the same single sum by inclusion-exclusion over empty rows. This is probably more straightforward.