How many ways can you sit $n-2$ people in a circular table of $n$ chairs? If I have a circular table of $n$ chairs, how many ways can I sit $n-2$ people?
For now my answer is finding the number of ways to sit $n$ people normally, which is $(n-1)!$.
Then divide it by $2!$ because the empty chairs are indistinguishable.
So my answer is $\frac{(n-2)!}{2!}$.
Edit: Meant to say $\frac{(n-1)!}{2!}$ sorry
Is this correct?
 A: First, remove two of the chairs, then sit the people down. You're sitting $n-2$ people in $n-2$ chairs, so there are $(n-3)!$ ways of doing it (up to rotational symmetry).
Now put back in the two chairs you removed. How many ways are there of doing this? There are $n-2$ spaces for the first chair, and $n-1$ spaces for the second; but as you note, the two chairs are indistinguishable. So there are $(n-1)(n-2)/2$ ways of putting the two chairs back in.
The end result is that there are $\frac{1}{2}(n-3)!\cdot(n-2)\cdot(n-1) = (n-1)!/2$ ways of seating $n-2$ people at a circular table with $n$ chairs.

Sanity checks with small numbers: Suppose $n=3$. Then you're seating $1$ person at the table, so up to rotational symmetry there's only one way; note that $(3-1)!/2=1$. Now suppose $n=4$. Then you're seating $2$ people at a table with $4$ chairs. Up to rotational symmetry there's no difference where the first person sits, and then there are $3$ choices for the second person, making a grand total of $3=(4-1)!/2$ ways.
