How many different necklaces can be made from seven black beads and three white beads?

Here is my approach so far:

Use a 10-gon as the representation of the necklace.

It follows that $10 \choose 3$ is the number of configuration of necklaces (3 being the number of white beads).

However, it is clear that we have overcounted (since we included rotations, e.g. 7 blacks followed by 3 whites and 3 whites followed by 7 blacks being counted separately). My approach is basically to look at the number of orbits in $D_5$.

|$D_5$| = 20; 1 + 9 rotations + 10 reflections.

My question is how I find |$G_x$|? I know $G_x$ is the stabilizer of a particular x, but I was wondering if I should find $G_x$ for a fixed x? What is X in this case? Do I just consider the stabilizers of each of the vertices of the 10-gon?

Thanks for your help.

• possible duplicate of Burnside's lemma - show that there are just five necklaces – Tacet Nov 30 '14 at 22:06
• @Tacet I am asking a specific question regarding stabilizers of $D_{10}$. The question is better detailed here. I do not believe this is a duplicate. – jstnchng Nov 30 '14 at 22:12
• I agree, there is better description. But if you have two same questions, maybe you should erase worst described? – Tacet Nov 30 '14 at 22:25
• This problem is best solved with Pólya's enumeration theorem, which follows from Burnside's lemma.. Are you ok with a solution using polya theory? – arindam mitra Dec 1 '14 at 8:58