In how many ways can pawns be placed on a $4 \times 4$ chessboard such that each row and each column contains exactly two pawns?
There are $\binom{4}{2}=6$ ways to place two pawns in the first row. Once the first row is placed, there are again 6 choices for the second row, and among these 6 choices, there is 1 where the first two rows are identical, 1 where they are opposite, and finally 4 choices such that the first two rows have exactly one pawn placement in common.

- First case: the first two rows are identical. Then the last two rows must be identical to the opposite of the first two, leaving no choice. This case thus represents 6 configurations.
- Second case: the first two rows are opposite. Then we have placed exactly 1 pawn per column, and so we can choose the third row in any way (thus in 6 ways) without risking placing more than two pawns in a column. Once the third row is placed, the last one is imposed. This case thus represents $6 \times 6=36$ configurations.
- Third case: the first two rows have exactly one pawn placement in common. Then there is one column with 2 pawns in which we can no longer place any pawns, and one column with 0 pawns in which we must necessarily place one pawn. The position of the second pawn can be chosen freely among the two remaining columns. There are thus 2 ways to choose the third row. Once the third row is placed, the last one is imposed. This case thus represents $6 \times 4 \times 2=48$ configurations. The number of possible placements is therefore $6+36+48=90$.