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?