TITLE: Number of White squares
QUESTION [5 upvotes]: In how many ways three white square can be selected on a $8 \times 8$ chessboard such that no two squares are in same row or column.
I am not able to reach on a conclusion for three squares. I have solved the problem for $2$ squares.
Please help me out and if possible provide a detailed explanation.

REPLY [3 votes]: Assume that $(0,0)$ is a white square. Call a white square even if it has even coordinates and odd, if it has odd coordinates; there are $4^2=16$ of each. The first square you can choose in $32$ ways; assume you pick an even one. The second square either is even, and there are $3^2=9$ left of these, or it is odd, and there are still $16$  of these. In the first case for the third choice $2^2=4$ even and $16$ odd squares remain, in the second case $9$ even and $9$ odd squares remain. Since the order in which the squares are selected is irrelevant there are
$${32\bigl(9\cdot(4+16)+16\cdot(9+9)\bigr)\over 6}=2496$$
possibilities in all.