10.3. Find the number of different ways to place 8 rooks in the cells of an 8x8 chessboard such that every cell of the board is under attack by at least one of them. The rooks can attack each other, a rook attacks all cells in the row and column it is in, including the cell it occupies.
Answer: $2 \cdot 8^{8}-8$ !

Solution. We will prove that the placement of 8 rooks satisfies the condition if and only if either there is a rook in each row, or there is a rook in each column, or both simultaneously. The positions of the rooks in the rows or columns are arbitrary. The sufficiency of the given condition is obvious.

If there is a column without a rook, then each of its 8 cells must be attacked by a rook standing on the same row. Therefore, if rooks are not in all columns, they must be in all rows.

The number of placements where there is a rook in each column is $8^{8}$, since for each of the 8 rooks, there are 8 independent ways to place it in one of the cells of the corresponding column. Similarly, there are $8^{8}$ placements where there is a rook in each row. However, placements where there is a rook in each column and each row, i.e., where the rooks do not attack each other, are counted twice, so the answer is obtained by subtracting the number of such placements, which is 8 !, from $8^{8}+8^{8}$. Therefore, the final answer is $2 \cdot 8^{8}-8$ !

Grading criteria. Proof that either there is a rook in each row or there is a rook in each column: 2 points. Correct calculation of the number of placements where there is a rook in each row (column): 1 point. Correct calculation of the number of placements where there is a rook in each column and each row: 1 point. 

Subtraction of the number of such placements from $8^{8}+8^{8}$: 3 points.