In how many ways can we choose a black square and a white square from a chessboard so that they are neither in the same row nor the same column?

From a chessboard $2$ small square are to be selected such that they are not in same row and same column?

Solution for $1$: 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.

Solution for $2$ This question was given by teacher and he said answer to this question will be $\frac{64*49}{2}$.

My doubt Why we did not divide by $2$ in first question? Why my teacher divided by $2$ in second question?


1 Answer 1


Both solutions are correct. You need to divide by $2$ in Sol 2 because $64*49$ counts the number of pairs of squares, but each $2$-element set of squares counts twice (first given by the pair $(a,b)$, then by the pair $(b,a)$).

In Sol 1 you do not overcount because you assume that the first square of a pair is black and the other is white so $(a,b)$ always determines the $2$-element set of squares uniquely.

  • $\begingroup$ It is a different problem (P2 is not the same as P1) So the answers are different too. $\endgroup$
    – markvs
    Jan 10 at 5:26
  • $\begingroup$ My mistake, I didn't read the posting closely enough. I have deleted my prior comment. $\endgroup$ Jan 10 at 5:28

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