The maximal amount of knights which can be places on a regular $8 \times 8$ chessboard so that no two take each other is $32$ (To see this just notice that a knight on a black square only attacks knights placed on white squares).
I was wondering how many ways there are to tile a regular chessboard with these $32$ knights so that no two take each other.
A generalisation of this question has already been posed here: In how many ways we can place $N$ mutually non-attacking knights on an $M \times M$ chessboard?. However, the answers given include a polynomial approach which cannot be used in the case of an $8 \times 8$ chessboard, because, as an answer states, a computer struggles to compute such a polynomial.
Even if it were quite easy for a computer to compute such a polynomial, in an Olympiad no such computational power is allowed. Hence, I was wondering if there was a pure combinatorics approach for this particular case.
An example of such a tiling is to place all the knights on all the black cells.