AP EAPCET 23 May 2025 Shift 1 Paper

We can choose three squares in a diagonal line parallel to BD in the △ABD. Clearly, three squares in △ABD and in a diagonal line parallel to BD can be chosen in
‌3C3+‌4C3+‌5C3+‌6C3+‌7C3+‌8C3 ways
Similarly, in △BCD the squares can be chosen parallel to BD in an equal number of ways.
Hence, the total number of ways in which three squares can be chosen in a diagonal line parallel to BD is
=2(‌3C3+‌4C3+‌5C3+‌6C3+‌7C3)+‌8C3
( ∵BD is common to both the triangles) Similarly, squares can be chosen in a diagonal line parallel to AC and hence the total number of favourable ways
=4(‌3C3+‌4C3+‌5C3+‌6C3+‌7C3)+2⋅‌8C3=392
Hence, the required probability
=‌

392

‌64C3

=‌

392×6

64×63×62

=‌

7

744