Circular permutation probability 
A circular table has $9$ chairs that $4$ people can sit down randomly. What is the probability for no two people sitting next to each other?

My current idea is to calculate the other probability, which is there are two people sitting next to each other, but I don't know enough conditions that I have to catch.
I only know that there would be $6!$ ways to arrange the other $7$ objects, the people-people pair can interchange places ($P_1P_2$ - $P_2P_1$) and I can choose which people that would form the pair ($_4C_2?$). Those conditions alone would give me $12 \times 6!$ arrangements, which is not likely (as I thought that two people sitting next to each other is more likely to happen than the other)
What did I miss?

This  problem can be  approached using  the Polya  Enumeration Theorem
(PET). Note that the four people  seated on the table create four gaps
between them that consist of empty  chairs and the empty chairs in all
four  gaps must  add up  to five,  where we  have  rotational symmetry
acting on the gaps.
Now the cycle index of $Z(C_4)$  is seen by enumeration to be (just
factor the four permutations)
$$Z(C_4) = \frac{1}{4} (a_1^4 + a_2^2 + 2a_4).$$
With the gaps possibly being empty we get
$$[z^5] Z(C_4)\left(\frac{1}{1-z}\right).$$
This works out to
$$\frac{1}{4} [z^5] \left(\frac{1}{(1-z)^4}
+ \frac{1}{(1-z^2)^2} + 2\frac{1}{1-z^4}
\right)$$
or
$$\frac{1}{4} {5+3\choose 3} = 14.$$
With the gaps containing at least one chair we get
$$[z^5] Z(C_4)\left(\frac{z}{1-z}\right)$$
which works out to
$$\frac{1}{4} [z^5] \left(\frac{z^4}{(1-z)^4}
+ \frac{z^4}{(1-z^2)^2} + 2\frac{z^4}{1-z^4}
\right)
\\ = \frac{1}{4} [z^1] \left(\frac{1}{(1-z)^4}
+ \frac{1}{(1-z^2)^2} + 2\frac{1}{1-z^4}
\right)$$
or $$ \frac{1}{4} \times {1+3\choose 3} = 1.$$
This finally yields for the desired probability
$$\frac{1}{14}.$$
Observe that by rotational symmetry if we must have at least one chair
in each gap  there is only one chair left to  distribute and these all
yield  the same  arrangement under  rotational symmetry,  so  there is
indeed just one such arrangement.

Addendum. For  the case of the people  being distinguishable (call
them  $A,B,C$ and  $D$  and let  $E$  represent gaps  as  in the  word
empty) we obtain for the total possibilities
$$[ABCDE^5] Z(C_4)\left((A+B+C+D)
\times (1+E+E^2+E^4+E^5+\cdots)\right)
\\ = [ABCDE^5] Z(C_4)\left((A+B+C+D) \frac{1}{1-E}\right).$$
Here the only contribution comes from the first term and we get
$$\frac{1}{4} [ABCDE^5] (A+B+C+D)^4 \frac{1}{(1-E)^4}
\\ = \frac{1}{4} {5+3\choose 3} [ABCD] (A+B+C+D)^4.$$
The case of gaps not being empty yields
$$\frac{1}{4} [ABCDE^5] (A+B+C+D)^4 \frac{E^4}{(1-E)^4}
\\ = \frac{1}{4} [ABCDE] (A+B+C+D)^4 \frac{1}{(1-E)^4}
\\ = \frac{1}{4} {1+3\choose 3} [ABCD] (A+B+C+D)^4.$$
We get for the probability
$$\frac{4\choose 3}{8\choose 3} = \frac{1}{14}.$$
Here  we  have attached  the  gap  next to  a  person  in a  clockwise
direction to the term for that person.
