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Math (Combinatorics)

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In how many ways can I place 5 different beads on a bracelet if flipping or rotating the bracelet does not change the order of the beads?

  • Math (Combinatorics) - ,

    The symmetry group here is D_5, see here:

    So, you have 10 different rotations and reflections (the identity, i.e. doing nothing is one of these operations).

    If you don't take into account this symmetry, you would have had 5!=120 possibilities. Applying different elements of D_5 always gives you a different configuration (this follows from the fact that you have 5 different beads), so the number of different configurations that can't be obtained from each other by applying a rotation or reflection is 120/10 = 12.

    If the beds are not all different, then it can happen that two different elements of D_5 applied to the same configuration yield the same result, dividing by 10 then won't give the correct result. In that case, you have to apply PĆ³lya enumeration theorem:

  • Math (Combinatorics) - ,

    Awesome, thanks!

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