How many ways are there to put 3 white rooks and 5 black rooks on an 8x8 chessboard, so none can attack each other? I've been trying to figure it out by using this solution as an example. I understand it pretty well, but can't wrap my mind around 3 white rooks and 5 black rooks instead of just 2 and 2.
 A: Place the white rooks first. There are several cases:


*

*All three white rooks on the same row or column. There are $16* {8\choose 3}$ ways of doing this, leaving $7*5 = 35$ squares free for the black rooks. So there are $16*{8\choose3}*{35\choose5}$ possibilities here.

*Two white rooks on the same row, two on the same column. (I.e. one rook shares a row with one of the other two and a column with the third). There are $8 * {8 \choose 2} * 14$ ways of doing this, and it leaves $64 - 8 - 7 - 7 - 6 = 36$ squares free for the black rooks, so there are $8 * {8 \choose 2} * 14*{36 \choose 5}$ possibilities here.

*Two white rooks on the same row or column, the remaining rook not sharing row or column with either of them. There are $16 * {8 \choose 2} * (64-8-7-7)$ ways of doing this, and leave $64-8-7-7-7-5 = 30$ squares free for the black rooks, so $16 * {8 \choose 2}*42*{30 \choose 5}$ possibilities here.

*All three white rooks on different rows and columns. There are $(64 * 49 * 36)/6$ possibilities, and leave $25$ squares free for the black rooks, so $64 * 49 * 6 * {25 \choose 5}$ possibilities here.
Total number of possibilities is 
$$16*{8\choose3}*{35\choose5} + 8 * {8 \choose 2} * 14*{36 \choose 5} + 16 * {8 \choose 2}*42*{30 \choose 5} + 64 * 49 * 6 * {25 \choose 5}$$
A: $8! {8 \choose{5}}$
You can just permute the numbers 1..8, and select 5 out of 8 rooks to be black.
