I found a similar problem and I studied the solution. But I still can not understand the solution exactly.

let S={necklaces of length 9 with 5 red and 4 green beads}.
So, $\left| S \right| =\frac { 9! }{ 5!4! } $
The group of symmetries, $ G={ D }_{ 9 }$.
$ \left| G \right| =\left| { D }_{ 9 } \right| =18$
Now, by the Burnside lemma, required answer is $$\frac { 1 }{ \left| G \right| } \times \sum _{ g\in G }^{ }{ fix(g) } $$ This part is understandable. I will ask you for help from now on, please.

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    $\begingroup$ So what is your question then? $\endgroup$ May 31 '17 at 7:21
  • $\begingroup$ With rotation but not turning over, I suspect there are $14$ necklaces with $5$ red and $4$ green: one with $4$ greens together, four with $3+1$ greens, two with $2+2$ greens, six with $2+1+1$ greens, and one with $1+1+1+1$ green. Allowing turning over as well as rotation, I think this becomes $10$ necklaces/bracelets (one, two, two, four, one of those types respectively) $\endgroup$
    – Henry
    May 31 '17 at 10:16

To do away with the problem of equivalence under rotation, fix one of the beads in its place, turning the problem into arranging others to its right, as on a shelf. However, I have a little doubt about what two necklaces are defined to be equivalent in your question.


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