There are 6 people at a party sitting at a round table with 6 seats: A

This question is different from the one posted above.

There are two versions and the answer would be different in the two cases.

Let me pick this version first:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, cannot sit next to D and F at the same time. (This means that A can sit next to D but not while F is on A's other side. Similarly, A can sit next to F too but not while D is on A's other side)

Total number of ways of arranging 6 people in a circle = 5! = 120

In how many of these 120 ways will A be between D and F?

We make DAF sit on three consecutive seats and make other 3 people sit in 3! ways.
or we make FAD sit of three consecutive seats and make other 3 people sit in 3! ways.
In all, we make A sit next to D and F simultaneously in 12 ways.

120 - 12 = 108 is the number of ways in which D and F are not sitting next to A at the same time.


The second version which seemed like the intended meaning of the original poster:
There are 6 people (say A, B, C, D, E and F). They have to sit around a circular table such that one of them, say A, can sit neither next to D nor next to F. (This means that A cannot sit next to D in any case and A cannot sit next to F in any case.)

Here, we say that A has to sit next to two of B, C and E.
Let's choose 2 of B, C and E in 3C2 = 3 ways. Let's arrange them around A in 2 ways (say we choose B and C. We could have BAC or CAB). We make these 3 sit on any 3 consecutive seats in 1 way. Number of ways of arranging these 3 people = 3*2 = 6

The rest of the 3 people can sit in 3! = 6 ways
Total number of ways in which A will sit neither next to D nor next to F = 6*6 = 36 ways