Question

12 persons are seated at a circular table. Find the probability that 3 particular persons always seated together.

Answer 1

Total probability$=(12-1)!=11!$

Desired probability$=(10-1)!=9!$

So, $P=\frac{9!\times 3!}{11!}=\frac{9!\times 3!}{11\times 10\times 9!}=\frac{3\times 2\times 1}{11\times 10}=3/55$ on simplification.