Number of White squares 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.
 A: First one can be placed in $8^2$ ways, given it second one can be placed in $7^2$ ways, given the first two third one can be placed in $6^2$ ways. Divide by $3!$ to account for the fact that the points could have been picked in any order. So total number of ways = $\frac{8^2.7^2.6^2}{3!} = 18816$
EDIT: Sorry did not realize it was a chessboard. Way to do it here would be to notice that once you choose parity of column you also choose parity of row of white square. So a square with odd row parity will never clash with square with even row parity. Now, either all three squares have same row parity or one is different from other two. In both cases we basically have disjoint instances of smaller problems like what I had earlier solved it as. In first case, number of ways = $\frac{4^2.3^2.2^2}{3!} = 96$. In second case, number of ways = $\frac{4^2.3^2}{2!} * \frac{4^2}{1!}= 72 * 16 = 1152$. So total = $1152+96=1248$. Now just notice that both of these cases can happen in two ways for two row parities, eg. for the first one it can be all odd row parities or all even row parities. So we need to double this to get $2 * 1248 = 2496$.
A: Let's first choose the three columns.  There are four cases,


*

*all columns even

*two columns even, one odd

*one column even, two odd

*all columns odd


It either of cases (1) and (4), we have four rows we could choose for the lowest numbered column, three for the next lowest numbered column, and two for the highest numbered column.  This gives
$$
2\cdot\binom{4}{3}\cdot(4\cdot3\cdot2)
$$
selections.
Now look at case (2).  We have four row choices for the lower even column, and three for the higher even column.  We have four row choices for the odd column.  Case (3) is similar, so we get
$$
2\binom{4}{2}\binom{4}{1}\cdot(4\cdot3\cdot4)
$$
selections.
A: 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.
A: Here's a computational solution. We can generate all such triples using this code in GAP:
S:=Filtered(Tuples([1..8],2),c->(c[1]+c[2]) mod 2=0);
T:=Filtered(Combinations(S,3),i->i[1][1]<>i[2][1] and i[1][1]<>i[3][1] and i[2][1]<>i[3][1] and i[1][2]<>i[2][2] and i[1][2]<>i[3][2] and i[2][2]<>i[3][2]);

This code gives 2496 triples of white squares, as per Wonder and Will Orrick's answers.
