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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.

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    $\begingroup$ possible duplicate of Burnside's lemma - show that there are just five necklaces $\endgroup$
    – Tacet
    Commented Nov 30, 2014 at 22:06
  • $\begingroup$ @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. $\endgroup$
    – jstnchng
    Commented Nov 30, 2014 at 22:12
  • $\begingroup$ I agree, there is better description. But if you have two same questions, maybe you should erase worst described? $\endgroup$
    – Tacet
    Commented Nov 30, 2014 at 22:25
  • $\begingroup$ 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? $\endgroup$ Commented Dec 1, 2014 at 8:58

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