Combinatorics: Selecting pebbles making up a 9-sided shape Find the number of ways to simultaneously select $3$ pebbles from $9$ pebbles that are arranged in a circle such that no two of the three selected pebbles are adjacent to each other, and one selection is the same as another rotation if the former can be rotated to form the latter.
My Solution: At first, we can pick any $9$ points. Then, any $6$ points. And at last, $4$ points. This turns out be $6\cdot4\cdot9$. However, we have over-counted. We divide this by $3$! because we do not care about the order. Lastly, we divide by $9$ to account for the last part of the problem. Answer: $4$
Is this correct? And is there an easier way using combinatorics?

We  may  also use  the  Polya  Enumeration Theorem.   Selecting  three
non-adjacent pebbles leaves a total of six non-selected ones that must
be placed in three available slots  with the cyclic group $C_3$ acting
on them and at least one pebble in every slot. The cycle index is
$$Z(C_3) = \frac{1}{3} a_1^3 + \frac{2}{3} a_3.$$
We then have
$$[z^6] Z\left(C_3; \frac{z}{1-z}\right)
= [z^6] \left(\frac{1}{3} \frac{z^3}{(1-z)^3}
+ \frac{2}{3} \frac{z^3}{1-z^3}  \right)
\\ =  \frac{1}{3} [z^3] \frac{1}{(1-z)^3}
+ \frac{2}{3} [z^3] \frac{1}{1-z^3}
= \frac{1}{3} {3+2\choose 2} + \frac{2}{3} [z^1] \frac{1}{1-z}  
\\ = \frac{1}{3} \times 10 + \frac{2}{3} = 4.$$
