Choosing 2 squares on an $8 \times 8$ ($64$ square) chessboard

Question

On an $8 \times 8$ ($64$ square) chessboard, how many ways can we choose pairs of squares such that each pair doesn't have the same colors? Each pair should consist of a white and black square.

Answer 1

Let
set of squares be $S$
set of white squares be $W$
set of black squares be $B$

then $W \cup B = S, \ W \cap B = \emptyset, \ |W| = |B| = \frac{|S|}{2}, \ |S| = 64 $

Consider $(s_1, s_2)$ ( $s_1, s_2 \in S$ ) be pair of squares chosen then
Ways of choosing pairs of $s_1$ as white and $s_2$ as black square
$ = |\{ (s_1, s_2) | s_1 \in W, s_2 \in B \}| \\ = |\{ s_1| s_1 \in W \} \times \{ s_2| s_2 \in B \}| \ [ \ \because W \cap B = \emptyset \ ] \\ = |\{ s_1| s_1 \in W \}| \times |\{ s_2| s_2 \in B \}| \\ = |W| \times |B| = \frac{|S|}{2} \times \frac{|S|}{2} = \frac{|S|^2}{4} $

Since order of choosing does not matter,
Ways of choosing pairs of squares of different color
= Ways of choosing pairs of $s_1$ as white and $s_2$ as black square
$ = \frac{|S|^2}{4} = \frac{64^2}{4} = 1024 $

Answer 2

Since there are 64 squares, 32 are black and 32 are white. You can choose a black square out of 32 in ${32}\choose{1}$ways and the same for the white squares. Then: ${32}\choose{1}$ • ${32}\choose{1}$$=32$•$32=1024$ ways of choosing a pair of squares (one white and the other black) out of 64.