Ways to seat r out of n people around a table

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Homework Statement



Find a formula for the number of ways to seat r of n people around a circular table, where seatings are considered the same if every person has the same two neighbors without regard to which side these neighbors are sitting on.

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The Attempt at a Solution



I understand the more basic form of this question; in order to seat n people, we have \frac{n!}{n}, since there are n! ways to seat n people and we divide out the n repetitions.

From there, we see that the ways to seat r of n people is simply \frac{P(n, r)}{r}. So that gives us the first part of the problem.

But I'm not really sure where to go from here. I'm thinking, in this specific problem, we shouldn't be dividing out the r people, since it's explicitly stated that the only seating arrangements that are considered the same are where the same person has the same two neighbors. So I'm thinking, I definitely have the numerator: P(n, r). And.. then I hit a wall.

Any help is appreciated.
 
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You'd still have to divide by r because if you shift everyone by one seat, they'll still have the same two neighbors. There's still another way to rearrange a permutation so that everyone has the same two neighbors. You need to account for that.
 
Okay, that makes sense. So now I have:

\frac{P(n,r)}{r}

What about multiplying r by 2, for the mirrored positions?

\frac{P(n,r)}{2r} = \frac{n!}{\frac{(n-r)!}{2r}}
 
Sounds good.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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