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?
 A: 
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?

To obtain a measure for the total space, we choose one person, lets name her Jo, and wherever Jo sits, count the ways to select three from the eight remaining chairs for the other people to sit relative to Jo.
To obtain a measure for the favored condition within that space, we count ways arrange the four people so everyone has an empty chair on their left.   That is: first reserve two seats for Jo, and then among the remaining seven seats, we count ways to select three adjacent pairs and one other empty seat.
A: Number the seats $0$ through $8$, and let one selected person sit at $0$.  Now, there are $\binom{8}{3} = 56$ different ways to select three seats for the remaining people, but only four of them—$\{2, 4, 6\}, \{2, 4, 7\}, \{2, 5, 7\}, \{3, 5, 7\}$—avoid people sitting next to each other.  (This can be seen from the fact that there is only one degree of freedom—an extra empty seat—to spare.  This empty seat can be placed in any of the four intervals between people.)  Therefore, the probability of such a configuration is $\frac{4}{56} = \frac{1}{14}$.
A: The complementary event for no two people sitting in consecutive seats is at least two people sitting in consecutive seats.  You tried to count the number of arrangements in which two people sit in consecutive seats, but there can also be two pairs of people in consecutive seats (either two disjoint pairs or two overlapping pairs of people in three consecutive seats) or three pairs of people in consecutive seats (if they sit in four consecutive seats).  To count those arrangements in which two or more pairs of people sit in consecutive seats, we need to apply the Inclusion-Exclusion Principle.  While it is possible to solve the problem that way, it is somewhat tricky.  It is easier to count those cases in which no two people sit in consecutive seats. 
We will solve the corresponding problem for people arranged in a row, then adjust our answer to account for the fact that we cannot place people at both ends of the row since those people would be sitting in adjacent seats when the ends of the row are joined to form a circle.  
We count linear arrangements in which no two people sit in consecutive seats:
Line up the four people in some order, say alphabetically.  Hand each of the four people a chair.  Line up five empty chairs in a row, leaving gaps between the chairs and spaces at the ends of the row in which chairs can be inserted.  This creates six spaces, four between successive chairs and two at the ends of the row.  To ensure that no two people sit in consecutive seats, the people must place their chairs in four of these six spaces.  The first person has six choices, the second person has five choices remaining, the third person has four choices remaining, and the fourth person has three choices remaining.  Hence, the number of ways four people can sit in nine seats arranged in a row so that no two of them occupy consecutive seats is 
$$6 \cdot 5 \cdot 4 \cdot 3$$
From these, we must exclude those seating arrangements in which people sit at both ends of the row.
We count linear arrangements in which no two people sit in consecutive seats and both ends of the row are occupied:
Line up the people as before.  Hand each person a chair.  Line up five empty chairs as before.  There are four ways to choose a person to sit at the left end of the row, which leaves three ways to choose a person to sit at the right end of the row.  To ensure that no two people sit in consecutive seats, the remaining two people must place their chairs in two of the four spaces between the five successive empty chairs.  They can do this in $4 \cdot 3$ ways.  Hence, the number of ways of seating four people in nine seats arranged in a row so that no two people sit in consecutive seats and both ends of the row are occupied is 
$$4 \cdot 3 \cdot 4 \cdot 3$$
Hence, the number of ways of seating four people in nine chairs arranged in a row so that no two people sit in consecutive seats and so that the two seats at the end of the row are not both occupied is 
$$6 \cdot 5 \cdot 4 \cdot 3 - 4 \cdot 3 \cdot 4 \cdot 3 = 216$$
Since we can transform a linear arrangement into a circular arrangement by joining the ends of the row, this is equal to the number of seating arrangements in a circle if the seats are labeled.  
The number of ways of seating four people in nine labeled chairs is 
$$9 \cdot 8 \cdot 7 \cdot 6 = 3024$$
Hence, the probability that no two people sit in consecutive seats if four people sit in nine labeled chairs arranged in a circle is 
$$\frac{216}{3024} = \frac{1}{14}$$ 
If the seats are not labeled, we must divide the number of seating arrangements by $9$ to account for rotational invariance.  Since we must do this in both the numerator and denominator, we arrive at the same probability.  
Note that we can express the numerator in the form 
$$6 \cdot 5 \cdot 4 \cdot 3 - 4 \cdot 3 \cdot 4 \cdot 3 = \frac{6!}{2!} - \frac{4!}{2!} \cdot 4 \cdot 3 = \frac{6!}{2!4!} \cdot 4! - \frac{4!}{2!2!} \cdot 4! = \left[\binom{6}{4} - \binom{4}{2}\right]4!$$
where the quantity inside the brackets represents the number of ways of selecting four of the nine seats in such a way that no two consecutive seats are chosen and the seats at the ends of the row are not both occupied, and the quantity outside the brackets represents the number of ways of arranging the people in the selected seats.
Note that we can express the denominator in the form 
$$9 \cdot 8 \cdot 7 \cdot 6 = \frac{9!}{5!} = \frac{9!}{5!4!} \cdot 4! = \binom{9}{4} \cdot 4!$$
where the quantity $\binom{9}{4}$ represents the number of ways of selecting four of the nine seats, and the quantity $4!$ represents the number of ways of arranging the people in the selected seats.
Hence, we can express the probability that four people sit at a round table containing nine seats so that no two consecutive seats are occupied as 
$$\frac{\left[\dbinom{6}{4} - \dbinom{4}{2}\right]4!}{\dbinom{9}{4}4!} = \frac{\dbinom{6}{4} - \dbinom{4}{2}}{\dbinom{9}{4}}$$
which means the probability depends only on which seats are occupied, not who sits in them.  
A: 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.
