Question: I am not able to take this question.My question is "Find the number of ways that can 12 persons take seats in a row of 20 fixed seats so that every person has exactly one neighbour"

My Attempt: I have started assuming that $x\_0$ be the number of vacant seats before the first person and $x\_1$ be the number between second and third and so on. Am i taking it right or wrong? Please guide if there is another way.

Answer: 
*A way using stars and bars*

* Form clumps of People and Empty chairs as below:

${\huge\boxed.}\; P P E\;{\huge\boxed.}\; P P E\;{\huge\boxed.}\; PPE\;{\huge\boxed.}\; PPE\;{\huge\boxed.}\; PPE\;{\huge\boxed.}\; P P\;{\huge\boxed.}$

* Place the $3$ remaining chairs in the boxes in $\binom{3+7-1}{7-1}$ ways
* Permute the individuals in $12!$ ways
Unless I'm missing something... there's only one way that is possible.

20 seats

I I I I I I I I I I I I I I I I I I I I

Each person has 1 neighbor.

I I x I I x I I x I I x I I x I I x I I

No other way to do it, since you can't offset the first or last seat and still have them have a neighbor.