How many ways are there to seat six people around a circular table where two seatings are considered the same when everyone has the same two neighbors without regard to whether they are right or left neighbors?
I have tried three different methods to solve this problem, all of which failed. I did find this solution on the internet, but I don't quite understand it. Could someone help me?
Solution:
We choose a distinguished person. Then, there are 5 people left. Hence there are 5! ways to seat these people. However, since we do not consider the diļ¬erence between right and left neighbors, each way we counted appears twice. Once for the counter clockwise arrangement and once for the clockwise arrangement. Hence, there are 5!/2 = 60 ways to seat the 6 people.