# Squares in chessboard that must not lie in the same row or same column

In how many ways is it possible to choose a white square and black square on a chessboard so that the squares must not lie in the same row or same column?

$a.)\ 56 \\ b.)\ 896 \\ c.)\ 60 \\ \color{green}{d.)\ 786}$

I did $\dbinom{64}{2}-\dbinom{32}{1} \dbinom{32}{1} =992$

But its none of the options .

I look for a short and simple way.

I have studied maths upto $12$th grade.

• 32*32 is the number of ways to choose a white square and a black square since there are 32 white squares and 32 black squares. Now how many ways are there to choose a white square and black square in the same row/column? Sep 9 '15 at 20:06

A chessboard contains $32$ white squares, so you have $32$ possible choices for the white square. Now in the same column or row of this square lie $8$ black square which you can't choose, leaving $32 - 8 = 24$ possible black squares to choose from. This yields a total of $32 \cdot 24 = 768$ possible choices.
• But in the corner lines there are $7$ black squares .
• But you choose a white square first, therefore you can't have this particular combination of a row and a column. Note that if you choose a white square, you have $4$ black squares in the same row and $4$ black squares in the same column as the white square. Sep 9 '15 at 20:46