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If n people are seated in a random manner in a row containing 2n seats, what is the probability that no two people will occupy adjacent seats?

Short Answer

Expert verified

The probability that no two people will occupy adjacent seats is \(\frac{{n{!^2} \times \left( {n + 1} \right)}}{{2n!}}\)

Step by step solution

01

Given information 

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

02

Compute the probability

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 2nth 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 \(\frac{{\left( {n + 1} \right){{\left( {n!} \right)}^2}}}{{2n!}}\).

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