# Another simple probability question

• March 31st 2012, 05:08 AM
alexmahone
Another simple probability question
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?
• March 31st 2012, 06:07 AM
Plato
Re: Another simple probability question
Quote:

Originally Posted by alexmahone
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? the answer given in the book is $\displaystyle P=\frac{(n+1)(n!)^2}{(2n)!}$.

Think of a string of n zeros and n ones.
There are $\frac{(2n)!}{(n!)^2}$ ways to arrange that string.
The n zeros create n+1 spaces to put the ones so that no two ones are together.
EXAMPLE: n=4, "_0_0_0_0_" that is five places.

Thus $\binom{n+1}{n}=n+1$ places seat the n people no two together.

$\frac{(n+1)}{\dfrac{(2n)!}{(n!)^2}}$ which is the book's answer.
• March 31st 2012, 06:35 AM
biffboy
Re: Another simple probability question
The first person can sit on any of 2n seats but the second person cant sit in next seat so doesnt have a choice of 2n-1 seats.
• March 31st 2012, 07:26 AM
alexmahone
Re: Another simple probability question
Quote:

Originally Posted by biffboy
The first person can sit on any of 2n seats but the second person cant sit in next seat so doesnt have a choice of 2n-1 seats.

Actually, I was counting the total number of arrangements.