Question

If I have a $4\times 4$ chess board, in how many ways can I place four distinct pawns on the board such that each column and row of the board contains no more than one pawn?

Answer 1

Each valid placement uses exactly one square in each row and each column, so the occupied squares form a permutation of the 4 columns across the 4 rows — there are 4! = 24 such square-sets. With four distinct pawns we can assign them to those 4 squares in 4! = 24 ways.

Total ways = 4!·4! = 24·24 = 576.