Using Burnside's Lemma find out:
How many different necklaces having $5$ beads can be formed using $3$ different kinds of beads, if we discount :
(a) Both flips and rotations?
(b) Rotations only?
(c) Flips only?
I don't know how to apply this lemma. Have you some solved exercises?
thanks :)
First, figure out what the symmetry group $G$ is for each part. For instance, for part (b) we are working with rotations - five of them - so we can say that $G\cong C_5$, the cyclic group of order $5$.
Then, for each element $g\in G$, figure out how many fixed points it has - i.e. the number of configurations which it preserves. Call each value $|X^g|$, or $\mathrm{Fix}(g)$.
Finally, tally these values together, and average i.e. divide out by the number of elements $|G|$.
As an example, with part (b) (the easiest), every nontrivial element $g\in G$ generates the whole group, so we know that if a necklace is fixed by $g$, then each bead must be the same color as every other bead. (Label the beads as numbers $1$-$5$ in order, then observe that $1,g1,g^21,\cdots$ lists out all of the numbers, so each bead must have the same color.) The number of necklaces with each bead the same color is equal to the number of colors, three, and the number of nontrivial elements of $G$ is four. Every necklace is fixed by the trivial element $e\in G$, and there are $3^5$ possible necklaces, so
$$|X/G|=\frac{1}{|G|}\sum_{g\in G}|X^g|=\frac{3^5+4\cdot3}{5}=51$$
is the number of necklaces modulo rotations.
