0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Hagen von Eitzen 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.