# Chess combinatoric problem

We are given eight rooks, five of which are red and three of which are blue. In how many ways can the eight rooks be placed on an 8-by-8 chessboard so that no two rooks can attack one another.

I don't know how to start this problem. Can anyone help?

• What does it mean for two rooks to "attack one another"? I guess a red rook and a blue rook can attack each other, but two red rooks can't attack each other because they're on the same team? – bof Sep 14 '17 at 1:47
• Do you mean that a blue rook cannot attack a red rook? EDIT: I didn't see bof's comment – Mr Pie Sep 14 '17 at 1:47
• There are no teams. The colours are there to differentiate the rooks so 5 of the rooks are the same and 3 of the rooks are the same. They can all attack one another. – Itsnhantransitive Sep 14 '17 at 1:49

If all $8$ rooks had the same color this would just give you a permutation matrix, of which there are $8!$. Given any of these permutation matrices we may simply choose $3$ of the rooks to change colors to blue, for a total of $8!\cdot\binom{8}{3}$

To expand on Thomas Grubb's great answer, here is the reason why if all rooks had the same color, there would be $8!$ possibilities:

You have $8$ rooks and $8$ rows. No two rooks can be placed in the same row, since they cannot attack each other. Therefore there must be one and only one rook on each row.

Now place the first rook in the first row. You have $8$ choices. Then place the second rook in the second row. You have $7$ choices; and so on.