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
$8$ empty seats left $(denoted~ by~ O)$

like O O O O O O O O

there are $7+2 = 9$ interspace to place the $6$ groups,

So i think the answer is $C^6\_9$ 

if the persons' order is alterable, 

then the answer is something like $C^6\_9$ \* $12!$