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

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.

• How many white squares are there? How many black? How many possible combinations are there then? – jvdhooft Feb 8 at 17:41
• Do you mean $8\times 8$? If the board is $10\times 10$ there are $100$ squares – pwerth Feb 8 at 17:42
• @pwerth Given that it's a chessboard, I took the liberty of changing it to $8 \times 8$. – jvdhooft Feb 8 at 17:43
• Sorry for the misunderstanding, I meant 10 X 10; I've edited the question. – williamcodes Feb 8 at 17:47
• @WilliamDarko Then please don't call it a chessboard, as a chessboard contains $8 \cdot 8 = 64$ squares. – jvdhooft Feb 8 at 17:51

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
$$= \frac{|S|^2}{4} = \frac{64^2}{4} = 1024$$

For each black square you can choose one of the 32 whites squares(32 possibilities), then, for the remaining 31 black and 31 white squares, you repeat the process (31 possibilities $$\times$$ the first 32 possibilities) ... . So the result should be $$32!$$.

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.