Question

$A$ and $B$ are two persons sitting in a circular arrangement with $8$ other persons. Find the probability that both $A$ and $B$ sit together.

Answer 1

Total number of people including $A$ and $B$ is $8+2=10$

As this is a case of circular permutations, ten people randomly be arranged around a circular table in $(n-1)!=(10–1)!=9!$ ways.

Since $2$ particular people are always together consider them as one person say $P$. Now, $P$ with other $8$, so $9$ can be arranged around the table in $8!$ ways. The two between themselves can arranged in $2!$ ways.

Hence the required probability is $\frac{8!2!}{9!}=\frac{8!2!}{9\times 8!}=\frac{2}{9}$