Two squares of sizé $$1\times1$$ are selected one after another from an $$8\times8$$ chessboard. The ‘probability that the two squares belong to different rows and different columns, is

First we'll select 1 square out of total 64 squares that means we can't select second square from that row and that column. Now we're left with 7 rows and 7 columns from which we can select another square. 

Hence, total square left = 7 * 7 = 49 [From which we can select another square].

Probability to select first square = $$\frac{64C1}{64C1} = 1$$

Probability to select first square = $$\frac{49C1}{63C1} = \frac{49}{63}$$

Required probability = $$1 * \frac{49}{63}$$

$$= \frac{7}{9}$$