# In using Polya's enumeration to count the number of necklaces made from two colours, how does placing a color to be equally spaced change the count?

By a Theorem, I am given that the number of orbits of some group $G$ acting on a set $B$ is $Z_G(|B|,\dots,|B|).$

I tried an example on $C_6$, the cyclic group of order $6$. Now, I got the cycle index $Z_{C_6}$ to be given by $\frac{1}{6}(2t_6+2t_3^2+t_2^3+t_1^6)$ which gives the number of orbits to be $\frac{1}{6}(2b+2b^2+b^3+b^6).$

Now, if I understand correctly, if I look at using this in terms of counting the number of necklaces I can make, I simply substitute the number of colours allowed for the beads I have for $b$ (say $b=2$ if I choose red and blue). But what happens when I add an extra restriction like only considering necklaces such that green beads are equally spaced amongst a total of $6$ beads? does my problem reduce to looking at $C_3$?

For equally spaced green beads among $6$ beads, you can have either $2$ green beads opposite to each other, or $3$ green beads in every other place.
Let's now put the green beads in and think about what the symmetry group (of rotations) of the necklace is. If there are $2$ green beads, a rotation must be of $C_2$ and in the case of $3$ green beads you have $C_3$. Count these two cases separately. You can now forget the green beads and only look at how the cyclic groups act on the other beads.