P and Q sit in a ring arrangement with 10 persons. What is the probability that P and Q will sit together?

Answer: Option A

n(S)= number of ways of sitting 12 persons at round table:
= (12 - 1)! = 11!
Since two persons will be always together, then number of persons:
= 10 + 1 = 11
So, 11 persons will be seated in (11 - 1)! = 10! ways at round table and 2 particular persons will be seated in 2! ways.
n(A) = The number of ways in which two persons always sit together = 10! × 2
$$\begin{array}{rl}& P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\\ & =\frac{10!\times 2!}{11!}\\ & =\frac{2}{11}\end{array}$$