Question

Total number of ways of selecting $3$ smallest squares on a normal chess board, so that they don't belong to the same row, same column or same diagonal line, is

• A. $18242$
• B. $18424$
• C. $18816$
• D. $18866$

Answer 1

The correct option is B $18424$

Total ways of selecting $3$ squares when they don't lie in the same row or column $=\frac{64\cdot 49\cdot 36}{3!}=18816$

Number of ways of selecting the squares when they lie in the same diagonal line
$=2[{}^8{C}_3+2({}^7{C}_3+{}^6{C}_3+{}^5{C}_3+{}^4{C}_3+{}^3{C}_3\left)\right]$
$=2[{}^8{C}_3+2{(}^8{C}_4)]$
$=392$

$\therefore$ Required number of ways
$=18816-392=18424$