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I'm working on a combinatorial problem and would appreciate some help.

Consider I have 9 balls consisting of three red balls, three green balls, and three blue balls. These balls are arranged in a circle (closed loop). Given that the loop is closed and the starting point does not matter, how many unique arrangements are possible?

I'm aware that in a linear arrangement, the number of unique permutations of the balls would be calculated using the multinomial coefficient:

9! / (3! * 3! * 3!)

However, because the balls are arranged in a circle, rotations of the same arrangement should be considered identical. I believe that this would involve dividing by the number of positions (9) to account for rotational symmetry, and possibly considering reflections if they are also counted as identical.

Could anyone provide a detailed explanation or formula for calculating the number of unique arrangements for this circular arrangement?

Thank you for your help!

vikash chaurasia
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    This problem can be thought as a class counting problem for a group action. I think https://math.stackexchange.com/questions/2016732/necklace-problem-with-burnsides-lemma already give some details. – Kolakoski54 Jun 15 '24 at 10:49
  • @Kolakoski54 Thank you for your kind help. I could get the answer. Any idea how to generate all such unique sequences? For example, let R,G, and B denote red, blue, and green balls. Sequences would be like RRRGGGBBB, RRRBBBGGG, etc. – vikash chaurasia Jun 15 '24 at 11:09
  • Dividing by $9$ will undercount the number of possible arrangements since there are only three distinguishable linear arrangements corresponding to the circular arrangement $BGRBGRBGR$. – N. F. Taussig Jun 15 '24 at 11:57

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