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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 know we have to apply division rule. I also know by rotating at six position we overcount by factor of 6 but that gives me answer of 120, while the correct answer is 60.

How to solve this?

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    $\begingroup$ $6$ people sit on a circular table in $(6-1)! = 120$ ways $\endgroup$ – Math Lover Oct 23 '20 at 8:24
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    $\begingroup$ Now if left or right neighbours do not matter, that is $1/2$. $\endgroup$ – Math Lover Oct 23 '20 at 8:25
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    $\begingroup$ Said differently: For each of your solutions, flip one clockwise-to-counterclockwise. They're equivalent. Hence divide by $2$. $\endgroup$ – David G. Stork Oct 23 '20 at 19:19
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All the guests can sit around the table in $6! = 720$ ways. Since two seatings are considered equal if they all have the same neighbor, the location of the first person, does not matter. Hence, a division by $6$. Also, since the left/right orientation is not an issue, we can 'flip' (place the first person at a seat, and instead of adding persons to the right, do it to the left) the seating and thus divide by $2$. This gives $60$ seatings in total.

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