$n\times n$ 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.

For odd $n$ there are no such rook placements on the black squares. Label the squares with ordered pairs from $(1,1)$ at the upper left to $(n,n)$ at the lower right. The white squares are now those $(x,y)$ for which $x+y=0 \mod 2$, while the black squares are those $(x,y)$ for which $x+y=1 \mod 2.$ The permutation caused by the rook placement on black squares now has the property that it maps an even to an odd, and an odd to an even.
Considering the cycle decomposition of the permutation, we can see that in each cycle we must alternate odd, even, odd, even etc. [note we do not have any one-cycles, i.e. the permutation has no fixed points, since for example a rook at $(5,5)$ is on a white square.] Also the length of each cycle must be even, since we cannot have e.g. the cycle $(a,b,c)$ where each of $a,b$, $b,c$, and $c,a$ have opposite parity. But then since each cycle has even length (and no fixed points), adding the lengths of the cycles gives an even number, which then cannot be $n$ as $n$ is odd.
