# Circular permutations such that everyone's nearest neighbours are the same

Suppose we want to find the number of ways to make $$n$$ people sit in a circle. All the arrangements in which everyone has the same nearest neighbours count as the same arrangement.

The total number of arranging $$n$$ people is $$n!$$

In any of these $$n!$$ arrangements, we can rotate the arrangement $$n$$ times and get the same arrangement.

How do we account for flips though? If everyone has a person sitting directly opposite to then, then we can flip the arrangement about that axis to get the same arrangement. We can also rotate the thing $$n$$ times following the flip to get the same arrangement. But flips can only be done if people are sitting directly opposite. There are no flips possible if, say, 5 people are sitting equally spaced apart in a circle.

• I don't understand the last sentence. If the people are sitting in the order 1-2-3-4-5 counterclockwise, why isn't seating then 1-2-3-4-5 clockwise count as a flip? Everybody has the same neighbors. – saulspatz Aug 4 at 3:04
• @saulspatz I didn't think about it that way. Thanks. So how do we account for these flips? I think it might be (n)!/2n. Because there are n rotations without the flip and n rotations after the flip. n+n=2n – Ryder Rude Aug 4 at 5:23
• Yes, that's right. – saulspatz Aug 4 at 6:06