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?
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 diﬀerence 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.