# 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

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?

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.