# Thread: Another simple probability question

1. ## Another simple probability question

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

My solution:

There are $\displaystyle 2n!$ favourable arrangements.

Total no. of arrangements:

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

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

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

2. ## Re: Another simple probability question

Originally Posted by alexmahone
If $\displaystyle n$ people are seated in a random manner in a row containing $\displaystyle 2n$ seats, what is the probability that no two people will occupy adjacent seats? the answer given in the book is $\displaystyle \displaystyle P=\frac{(n+1)(n!)^2}{(2n)!}$.
Think of a string of n zeros and n ones.
There are $\displaystyle \frac{(2n)!}{(n!)^2}$ 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 $\displaystyle \binom{n+1}{n}=n+1$ places seat the n people no two together.

$\displaystyle \frac{(n+1)}{\dfrac{(2n)!}{(n!)^2}}$ which is the book's answer.

3. ## 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.

4. ## Re: Another simple probability question

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.