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?

Re: Another simple probability question

Quote:

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.

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.

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.