TG EAMCET 2 May 2025 Shift 1 Paper

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
⇒‌‌‌

96

41664

=‌

1

434