Question

If 3 squares are selected at random on a chessboard having 8x8 squares, then the probability that they will be in a diagonal line is

• A. $\frac{\{{}^8{C}_3+2{(}^7{C}_3{+}^6{C}_3{+}^5{C}_3{+}^4{C}_3{+}^3{C}_3)\}}{{}^{64}{C}_3}$
• B. $\frac{2\{{}^8{c}_3+{(}^7{C}_3{+}^6{C}_3{+}^5{C}_3{+}^4{C}_3{+}^3{C}_3)\}}{{}^{64}{C}_3}$
• C. $\frac{2\{{}^8{C}_3+2{(}^7{C}_3{+}^6{C}_3{+}^5{C}_3{+}^4{C}_3{+}^3{C}_3)\}}{{}^{64}{C}_3}$
• D. $\frac{\{{}^8{C}_3+{(}^7{C}_3{+}^6{C}_3{+}^5{C}_3{+}^4{C}_3{+}^3{C}_3)\}}{{}^{64}{C}_3}$

Answer 1

The correct option is C $\frac{2\{{}^8{C}_3+2{(}^7{C}_3{+}^6{C}_3{+}^5{C}_3{+}^4{C}_3{+}^3{C}_3)\}}{{}^{64}{C}_3}$

The main diagonal has $8$ squares and there are $2$ such diagonals.
There are other minor diagonals having $7,6,5,4,3$ squares and there are $4$ such diagonals each.
We need to select $3$ squares from the diagonals.
This can be done in $\left({(}_3^{8}C)\ast 2\right)+\left({(}_3^{7}C{+}_3^{6}C{+}_3^{5}C{+}_3^{4}C{+}_3^{3}C\right)\ast 4)$ ways.

Hence, probability = $\frac{2\ast {(}_3^{8}C)+(2\ast {(}_3^{7}C{+}_3^{6}C{+}_3^{5}C{+}_3^{4}C{+}_3^{3}C\left)\right)}{{}_3^{64}C}$