Number of non-equivalent necklaces A necklace is made up of 12 beads in a circular loop. 3 are green and 9 are yellow. How many non-equivalent necklaces can be made?
I have to use Burnside's Counting Lemma in this question.
 A: I assume the group that establishes the admissible symmetries is simply the cyclic group $G = \langle a \rangle$ of order $12$, where $a$ is a rotation of $\dfrac{2 \pi}{12}$. (Addendum: from a comment of OP, it seems the group is the dihedral group $D$ of order $24$, which makes the situation slightly different, see below.)
Consider what are the possible orders of the elements of $G$, and how many there are of a given order. For each of these elements, call it $b$, consider whether there are necklaces fixed by $b$, and in case how many. 
For instance, if $b$ is an element of order $4$, take a green bead, and trace it as you apply the four distinct powers of $b$. You see that this would imply that there are $4$ green beads, which is not the case. If you consider an element $b$ of order $3$ (there are two of these in $G$), you will see that the situation is different.
Addendum for the dihedral group $D$. We have to consider the reflections, that is, the elements of $D \setminus G$. If you consider a reflection $\sigma$, it will fix a necklace if and only if its axis of symmetry goes through two opposite beads, and one of them is green, plus the other two green beads are symmetric with respect to $\sigma$. So for each of these $6$ reflections (half of the total reflections, as the other 6 have the axis going through the midpoints of two opposite sides) there are $2 \cdot (12 - 2)/2 = 10$ beads fixed by it.
And then of course the identity fixes all the possible necklaces: how many are there?
Let me know whether you are able to do the counting with these hints.
