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#### Alexmahone

##### Active member

- Jan 26, 2012

- 268

__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?