Example 1 In an $8 \times 8$ chessboard, how many ways are there to select 56 squares such that: all the black squares are selected, and each row and each column has exactly seven squares selected? ${ }^{[1]}$
(2014, Irish Mathematical Olympiad)
The problem is equivalent to selecting eight white squares on the chessboard, with exactly one square selected from each row and each column.

All the white squares on the chessboard form a $4 \times 4$ submatrix at the intersections of rows $1, 3, 5, 7$ and columns $1, 3, 5, 7$, and another $4 \times 4$ submatrix at the intersections of rows $2, 4, 6, 8$ and columns $2, 4, 6, 8$. Since there are 4! ways to choose four white squares from each submatrix such that no two squares are in the same row or column, the total number of ways to make the selection is $(4!)^2 = 576$.

[Note] In this problem, considering the "complement" of the selected squares is an important step in simplifying and visualizing a complex proposition.