Necklace combinatorics problem

Question

If $n$ is the number of necklaces which can be formed using $17$ identical pearls and two identical diamonds and similarly $m$ is the number of necklaces which can be formed using $17$ identical pearls and $2$ different diamonds, then the value of $m$ and $n$ is?

Answer 1

Use Burnside's Lemma.

If $G$ is a finite group of permutations on a set $S$, then the number of orbits of $G$ on $S$ is \begin{align} \frac{1}{|G|}\sum_{g \in G}|\operatorname{fix}(g)|. \end{align}

Observe there are $\binom{19}{2}=171$ possible designs (of course, some of them are equivalent by rotation and reflection). Let $G = D_{38}$ act on the $171$ necklace designs. Observe we see that \begin{matrix} \text{element} & \text{Number of Designs Fixed by Element}\\ e & 171\\ r^k, k=1,\ldots, 18 & 0\\ f & 9\\ fr^k, k=1,\ldots, 18 & 9 \end{matrix} Hence by Burnside's lemma, we see that \begin{align} \frac{1}{38}(171+9+9*18) = 9. \end{align}

Indeed, one possible configuration is that the two diamonds are $1$ apart on one side and $16$ apart on the other side or $3$ apart on one side and $14$ apart on the other side...until $17$ apart on one side and $0$ apart on the other side. Hence $9$ configurations.