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?

By your method, there are $64$ ways to place the first rook.  This attacks $15$ squares, leaving $49$ ways to place the 2nd rook.  This attacks 13 squares not attacked by the first rook, leaving $36$ was to place the third rook, and so on...
$64 \times 49 \times 36 \times 25 \times 16 \times 9\times 4 \times 1 = (8!)^2$
But since every rook is identical and it doesn't matter what order we place them, we must divide by the number of orders in which they can be placed, which is $8$ choices for the first rook, $7$ choices for the 2nd rook, $6$ choices for the 3rd rook and so on, so we will divide by $8!$ to convert the permutations into unique combinations:
$$\frac{(8!)^2}{8!}=8!$$
