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Another simple propability question

Alexmahone

Active member
Jan 26, 2012
268
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?

My solution:

There are $2n!$ favourable arrangements.

Total no. of arrangements:

The 1st person can sit on any of the $2n$ seats, the 2nd person can sit on any of the other $2n-1$ seats and so on. So, the number of ways to seat $n$ people on $2n$ seats is $\displaystyle 2n(2n-1)\cdots (n+1)=\frac{(2n)!}{n!}$.

$\displaystyle P=\frac{2(n!)^2}{(2n)!}$

But the answer given in the book is $\displaystyle P=\frac{(n+1)(n!)^2}{(2n)!}$. Where have I gone wrong?
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
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?

My solution:

There are $2n!$ favourable arrangements.

Total no. of arrangements:

The 1st person can sit on any of the $2n$ seats, the 2nd person can sit on any of the other $2n-1$ seats and so on. So, the number of ways to seat $n$ people on $2n$ seats is $\displaystyle 2n(2n-1)\cdots (n+1)=\frac{(2n)!}{n!}$.

$\displaystyle P=\frac{2(n!)^2}{(2n)!}$

But the answer given in the book is $\displaystyle P=\frac{(n+1)(n!)^2}{(2n)!}$. Where have I gone wrong?
The number of ways to place $n$ people on $2n$ seats such that no two sit adjacent to each other is not $2(n!)$. Its equal to $(n+1)(n!)$.
Take $n=2$ and list out the possible ways to seat 2 people on 4 seats. You will realize your mistake.