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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 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.

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It would help if you said more specifically which part of the solution you don't quite understand. –  joriki Apr 21 '13 at 18:28

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They start out by saying that it is easiest to think about the problem by fixing one person's spot, and considering arranging everyone else around that chosen person. This will not affect anything; after all, no matter how you arrange the people, you can always speak about everyone's position relative to one chosen person!

They then notice that the number of ways to arrange $n$ distinct object in a row is $n!$ which here gives us $5!=120.$ That is a fact, and for more information on arranging objects (also known as permutations) see here; this is a peek into the realm of math called Combinatorics.

However, $n!$ overcounts (more precisely, doubles) the number that we are looking for, since that formula distinguishes with regard to order. That is to say, it counts both of the possibilities of the right/left neighbors (i.e., Person A on the left and Person B on the right, or vice-versa). But we do not distinguish between these two possibilities! So it turns out that $5!$ doubles our desired number. So all that remains is "to get rid of" the overcounted arrangements by dividing out by two, and so we get $5!/2=60$ as our answer.

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