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.

  • $\begingroup$ 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? $\endgroup$
    – jschnei
    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.

  • $\begingroup$ But in the corner lines there are $7$ black squares . $\endgroup$
    – R K
    Sep 9 '15 at 20:10
  • $\begingroup$ @RK Count again. Recall that we consider rows and columns intersecting in white $\endgroup$ Sep 9 '15 at 20:11
  • $\begingroup$ See this prntscr.com/8eid4a $\endgroup$
    – R K
    Sep 9 '15 at 20:16
  • $\begingroup$ 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. $\endgroup$
    – Dominik
    Sep 9 '15 at 20:46

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