Problem: How many ways can 8 mutually non-attacking rooks be placed on the $9 \times 9$ chessboard (diagram omitted) so that all 8 rooks are on squares of the same color? (the four edge tiles are black) Solution: Let's do some case work! * Case 1: The 8 rooks are placed on white tiles. In this case, notice that, while there are $5$ rows with $4$ white tiles, we cannot choose all of them to put rooks on (by pigeonhole). In fact, we *must* place a rook in all $4$ rows with $5$ white tiles and then choose $4$ rows from the $5$ rows with $4$ white tiles to put the rest of the rooks in. This case contributes $\binom{5}{4}(5 \cdot 4 \cdot 3 \cdot 2)4!$ since for each way of choosing the $4$ rows from the $5$ with $4$ white tiles, we have $4!$ ways of arranging them, and then there are $5 \cdot 4 \cdot 3 \cdot 2$ ways of arranging the rooks on the $4$ rows with $5$ white tiles. * Case 2: The 8 rooks are placed on black tiles. We have two sub-cases * Case 2.1: We have $5$ rooks occupy all $5$ rows with $5$ black tiles and choose $3$ rows from the $4$ containing $4$ black tiles for the rest to go on. In this case, we can choose the $3$ rows with $4$ black tiles by $\binom{4}{3}$. For each choice, we have $4 \cdot 3 \cdot 2$ ways of putting the rooks on these rows. In the remaining rows with $5$ black tiles, there are $5!$ ways of ordering the rooks. In total, this case contributes $\binom{4}{3} (4 \cdot 3 \cdot 2) 5!$ * Case 2.2: We have $4$ rooks occupy $4$ rows with $5$ black tiles and have the rest of the rooks occupy all $4$ rows containing $4$ black tiles. We choose $4$ of the $5$ rows that have $5$ black tiles with $\binom{5}{4}$ and then order the $4$ rooks on these rows by $(5 \cdot 4 \cdot 3 \cdot 2)$. Then, on the $4$ rows with $4$ black tiles, there are $4!$ ways to arrange them. This case contributes $\binom{5}{4} (5 \cdot 4 \cdot 3 \cdot 2) 4!$ All together, we get that $\binom{5}{4}(5 \cdot 4 \cdot 3 \cdot 2)4! + \binom{4}{3} (4 \cdot 3 \cdot 2) 5! + \binom{5}{4} (5 \cdot 4 \cdot 3 \cdot 2) 4! = 40320$.