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.

The white squares on the chessboard are formed by the intersections of rows $1, 3, 5, 7$ and columns $1, 3, 5, 7$, resulting in a $4 \times 4$ submatrix, as well as the intersections of rows $2, 4, 6, 8$ and columns $2, 4, 6, 8$, resulting in another $4 \times 4$ submatrix. Since there are $4!$ ways to choose four white squares from each submatrix such that they are in different rows and columns, the total number of ways to make the selection is $(4!)^{2}=576$.