$n\times n$ board, non-challenging rooks Consider an $n \times n$ board in which the squares are colored black and white in
the usual chequered fashion and with at least one white corner square.
(i) In how many ways can $n$ non-challenging rooks be placed on the white
squares?
(ii) In how many ways can $n$ non-challenging rooks be placed on the black
squares?
I've tried some combinations with shading z certain square and deleting the row and the column in which it was and then using recurrence, but it doesn't work.

We are looking at all permutations of $n$ that are (white squares case) parity-perserving, or (black square case) parity-inversing. If $n$ is even the black squares case is equivalent to the white square case (by a vertical reflection for instance), and if $n$ is odd the black square case has no solutions (the odd numbers $1,3,\ldots,n$ must map bijectively to fewer even numbers $2,4,\ldots,n-1$).
So it remains to do the white square case. But now we must permute the odd and the even numbers independently among themselves, for $\lfloor\frac n2\rfloor!\times\lceil\frac n2\rceil!$ solutions.
