# How many different necklaces can be make from 5 red beads and 4 green beads?

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.

• So what is your question then? May 31 '17 at 7:21
• 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) May 31 '17 at 10:16