Question

Two squares of a chess board having 8x8 squares are selected at random. The probability that they have exactly one corner in common is

• A. $\frac{\{7+2(6+5+4+3+2+1\left)\right\}}{{}^{64}{C}_2}$
• B. $\frac{2\{7+(6+5+4+3+2+1\left)\right\}}{{}^{64}{C}_2}$
• C. $\frac{2\{7+2(6+5+4+3+2+1\left)\right\}}{{}^{64}{C}_2}$
• D. $\frac{\{7+(6+5+4+3+2+1\left)\right\}}{{}^{64}{C}_2}$

Answer 1

The correct option is C $\frac{2\{7+2(6+5+4+3+2+1\left)\right\}}{{}^{64}{C}_2}$

In a 8*8 square, there are 7*7 squares on the inside which can have a corner common with another one. For each such corner, the squares may be selected in 2 ways (along either diagonal direction).
Hence, probability $=\frac{2\ast (7\ast 7)}{{}^{64}{C}_2}=\frac{98}{{}^{64}{C}_2}$
This matches with option (c).
(Note: in the options, they have selected 2 consecutive squares diagonally. There are 2 major diagonals with 8 squares and 4 minor diagonals with $7,6,5,4,3,2$ squares. Hence, the form given)