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Hello could anyone tell me if I'm on the right track? I'm a little confused with circular arrangements

This is the question : How many ways can 8 persons sit in a circle with 10 seats such that there is always exactly one person between the empty two seats?

is the answer $\frac{9!}{2}$•8 ?

My reasoning is to treat the person in the middle and the two empty chairs as one group, so then there are $\frac{9!}{2}$ possible permutations for that arrangement since the two empty chairs are identical in each arrangement. Then the individual in the middle could be any of the 8 people, resulting in $\frac{9!}{2}$•8

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  • $\begingroup$ Suppose I was to extend your logic to $2$ persons and $4$ chairs in which case we know there is only one way to seat them. You would get $\frac {3!}2\times2=6$ ways, so there is clearly something amiss. Have you tried counting the number of groups you actually have? They are not $10$. $\endgroup$ Nov 1 '20 at 11:17
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By convention, in a circular permutation, only the relative order of the objects matters. Therefore, we can take the person who sits between the two empty seats as our reference point.

There are eight ways to choose the person who sits between two empty seats. As we proceed clockwise around the table from that person, there are $7!$ ways to seat the remaining seven people in the remaining seven seats, that is, the seats we have not reserved for the person who sits between two empty seats and the two empty seats. Hence, there are $$8 \cdot 7! = 8!$$ such seating arrangements.

The problem with your approach is that the two empty seats must be separated by exactly one seat. Therefore, you do not have $10$ objects to permute since the location of the solitary person also determines the positions of the empty seats.

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