Number of ways to arrange $8$ rooks on a chessboard Find the number of ways to arrange $8$ rooks on a chessboard such that no two of them attack other?
I was thinking it would be $64 \times 49 \times 36 \times 25 \times 16 \times 9\times 4 \times 1$, because after selecting 1 rook, the only squares left for selecting the next rook would be $15, 13, 11, 9 , 7 , 5$ and $3$ less than when selecting for the current rook?
Can someone explain?

Obviously you have to place a rook on the each of the $8$ columns. Now for the first column we have $8$ options. For the second we have $7$, as one of the rows is already taken by the previous rook and so on. Eventually you will get that the number of ways to place them is $8! = 40320$.
To see why your reasoning is false note that the rooks aren't distinguishable. Using your idea you count the way of placing $8$ distinguishable rooks on the chessboard. Hence because the rooks can permutate in $8!$ ways between themselves, you need to divide by $8!$ and you will get exactly the same answer.
Anyway if you want to read more about this problem you may want to look here.
