Number of ways that can 12 persons take seats in a row of 20 fixed seats so that every person has exactly one neighbour 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.
 A: There will be $8$ seats not taken. Write down $8$ stars, like this, with a gap between them.
$$\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast\qquad\ast$$
These determine $9$ gaps, of which $7$ are interstar gaps, and $2$ are endgaps. We need to choose $6$ of these gaps, and slip a couple of chairs in the chosen gaps. The choosing can be done in $\binom{9}{6}$ ways. 
If we are interested in who sits where, multiply by $12!$.
Added: The approach begun in the OP will work. We end up with $7$ variables $x_0$ to $x_6$ that have sum $8$. Close to a basic Stars and Bars problem, except that the "end" variables can be $0$, while the variables $x_1$ to $x_5$ must be $\ge 1$. We can break into cases, though there is a trick that gets us $\binom{9}{6}$ directly.
A: 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 
A: 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.
A: $8$ 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 $C^6_9$ * $12!$   
