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?
 A: 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
$$[z^6] Z\left(C_3; \frac{z}{1-z}\right)
= [z^6] \left(\frac{1}{3} \frac{z^3}{(1-z)^3}
+ \frac{2}{3} \frac{z^3}{1-z^3}  \right)
\\ =  \frac{1}{3} [z^3] \frac{1}{(1-z)^3}
+ \frac{2}{3} [z^3] \frac{1}{1-z^3}
= \frac{1}{3} {3+2\choose 2} + \frac{2}{3} [z^1] \frac{1}{1-z}  
\\ = \frac{1}{3} \times 10 + \frac{2}{3} = 4.$$
A: You have obtained the correct answer, but by accident.  As pointed out in the comments by John Hughes and G Cab, it is not the case that after picking the first two points you always have four choices for the third point: in some cases you have only three choices.  This reduces the number of configurations that you have prior to accounting for equivalences.  On the other hand, the size of the equivalence classes is not always $54=6\times9$: for configurations that look like an equilateral triangle, there are only three rotated versions of the configuration, not nine.  It seems that by lucky coincidence, these two overestimates cancel each other out.  I don't see any deep reason for this, and you can check that your method doesn't work when the number of pebbles is changed from 9 to 8 or 10 (or to 6 or 12, which are similar to 9 in that they also have equilateral triangle configurations).
A method that does work is to use the methods given here or here to compute the number of ways of selecting three pebbles, no two adjacent.  Using Brian Scott's method, you get
$$
\binom{9}{3}-9\cdot(9-4)-9=30.
$$
Now the three configurations that look like equilateral triangles are equivalent under rotation and the 27 remaining configurations come in equivalence classes of size 9.  This gives $1+\frac{27}{9}=4$ equivalence classes.
You can also account for the symmetry using Burnside's lemma: of the nine elements of the rotation group, the identity fixes all $30$ configurations and clockwise rotation by $3$ or $6$ spaces fixes the three equilateral triangle configurations. Hence the number of orbits is
$$
\frac{1}{9}(30+2⋅3)=4.
$$
