91Ó°ÊÓ

n people are sitting in a row of 2n seats in a random manner.

The total number of ways to select seats from 2n is \(\left( {^{2n}{C_n}} \right)\).

Define a new event that none of the two people occupied adjacent seats, which is feasible when the even seats are left empty and the odds are occupied.

In this way the last seat, the 2n^th would be left empty.

It implies, there are\(\left( {{\bf{n + 1}}} \right)\)possibilities for the arrangement,

Thus, the probability of event A, that no two peoples will occupy adjacent seats,

\(\begin{aligned}{}{\rm P}\left( A \right) = \frac{{\left( {n + 1} \right)}}{{^{2n}{C_n}}}\\ = \frac{{n + 1}}{{\frac{{2n!}}{{\left( {2n - n} \right)!n!}}}}\\ = \frac{{\left( {n + 1} \right){{\left( {n!} \right)}^2}}}{{2n!}}\end{aligned}\)

Thus, the required probability is \(\frac{{\left( {n + 1} \right){{\left( {n!} \right)}^2}}}{{2n!}}\).