Total number of ways to choose any 3 distinct squares =‌64C3 ⇒‌‌‌
64!
3!61!
=‌
64×63×62
6
=41664 Since, each row has 8 squares. In each row, the number of ways to choose 3 adjacent squares =8−3+1=6 So, total for 8 rows =8×6=48 Similarly, total for 8 columns =8×6=48 ∴ Total favourable outcomes ⇒‌‌48+48=96 Hence, required probability ⇒‌‌‌