TS EAMCET 9 May 2024 Shift 2 Solved Paper

Given that the number of circular permutation of 9 distinct things taken 5 at a time is n1.
∴‌‌n1‌=‌9C5(5−1)!
‌=‌

9!

5!(9−5)!

×4!=9×8×7×6
Also, given that the number of linear permutation of 8 distinct things taken 4 at a time is n2.
∴n2=‌8C4(4!)‌=‌

8!

4!(8−4)!

×4!
‌=8×7×6×5
Now, ‌

n1

n2

=‌

9×8×7×6

8×7×6×5

=‌

9

5