In how many ways can five indistinguishable rooks be placed on an 8-by-8 chessboard so that no rook can attack another and neither the first row nor the first column is empty?
I started out by calculating the total number of ways of placing $5$ rooks on $8\times8$ board so that no rook can attack another which is $\binom{8}{5}5!$. Then I calculated the number of ways in which 5 rooks can be placed by leaving both first row and first column empty, which then corresponds to placing 5 rooks on $7\times7$ board (because first row and column are out of picture), this can be done in $\binom{7}{5}5!$ ways. Then I subtracted this from total, giving me $4200$ ways. But I think this method is not correct as I am just subtracting the case of leaving both first row and first column empty, which should not lead to just getting both first row and first column non-empty.
My another approach is this - Calculate the ways where there is a combination where one rook gets placed in intersection of first row and first column or, where one rook gets placed in first row and another rook gets placed in first column.
For the first part, there is only one way to place a rook in the intersection of first row and first column, and rest 4 rooks are now to be placed on $7\times7$ board which can be done in $\binom{7}{4}4!$ ways. For the second part, take anyone of the squares in the first row (not intersecting with first column) and place a rook on that. This can be done in $7$ ways. Then we have remaining 7 rows and 7 columns (including first column). Again take one rook and place it in first column, which can be done in $7$ ways. Now 6 rows and 6 columns are remaining along with 3 rooks, which can be placed in $\binom{6}{3}3!$ ways. So total ways I get is $\binom{7}{4}4! + 7\times7\times\binom{6}{3}3! = 6720$.
Please tell me which of my approach is correct and if correct, then have I done the correct calculation or missed something.