# Circular permutation - Arranging 4 persons around a circular table where 8 seats are there. (cond.)

Suppose 4 persons A,B,C and D sit around a round table with 8 seats. Rotation by 8,16,24,... seats defines same arrangement and other rotations gives different arrangements.

If seats are identical, there are 7*6*5 arrangements as clarified here. After sturdying such questions,the following doubt came to my mind.

Suppose persons and seats are identical. If so, what is the required number of ways that these four people can be seated at the round table.?

How to approach such problem?

I guess this may not be simple as my previous question cited. I think, if persons and sets are identical, different arrangements may get identified only based on the empty seats between persons. How to approach this problem?

• How can persons be identical ? – true blue anil Mar 12 '16 at 14:58
• @true blue just took it as en example. Suppose 4 identical objects in place of 4 persons if that is an issue – Kiran Mar 12 '16 at 14:59
• Ok, consider it as 4 identical objects in 8 identical places around a circle – Kiran Mar 12 '16 at 15:01
• Yes, it then becomes a great deal more complicated ! Look up necklaces with beads of two colors (one for filled seats, and one for empty seats). To start you off, math.stackexchange.com/questions/1015418/… – true blue anil Mar 12 '16 at 15:18
• According to the link that you provided, $7\cdot6\cdot5$ is NOT the answer (OP thought it was the answer but got a different one). – barak manos Mar 13 '16 at 11:30

This problem is similar to the problem of counting the different necklaces with 4 whites beads and 4 black beads. It's possible to solve that using a group acting on a set. The set is all the possibilities of coloring a fixed necklace with 4 black and 4 white beads. The group is the group of the rotations of an octagon ($\{k\frac{\pi}{4} \pmod{2\pi} \mid k\in \mathbb{Z}\}$, $+$). Then, by counting the size of each orbit, it's possible to deduce the number of fundamentally different colorations of the necklace. If you have already heard of groups and groups acting on sets, I can tell you more about this method.