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Number of different necklaces using $m$ red and $n$ white pebbles

I don't understand high level maths. Please try to do simple$\ldots$ I tried to hunt down the pattern but couldn't complete as I have my exam.

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marked as duplicate by Douglas S. Stones, Erick Wong, Asaf Karagila, Austin Mohr, Fabian Dec 31 '12 at 8:28

This question was marked as an exact duplicate of an existing question.

There is no reason to cripple your post below twitter message length or the like. So please use complete words. – Hagen von Eitzen Oct 14 '12 at 16:10
Are you familiar with Burnside's lemma? The only way I see to do it is with that, but I don't know if you know Burnside's lemma. – only Oct 14 '12 at 16:29
Please clarify whether you mean "necklace" in the technical sense of considering arrangements related by a rotation as equivalent, as mentioned in sdcvvc's comment under Jen-Ya's answer, or in the everyday sense where a necklace has a distinguished spot where it opens and closes, so that rotations do matter. – joriki Oct 14 '12 at 18:22

[Edit] Sorry, I was not aware that necklaces are considered equivalent up to rotation. Thanks to sdcvvc for correcting

This solution is only right for situations, where rotations are not equivalent:

(n+m)! / (n! * m!)

You have n+m Beads and (n+m)! ways to order them. But you have to ignore the order of black beads and the order of white beads. There are n! permutations (possibilities to order) of black beads and m! permutations of white beads. So you have (n+m)! / (n! * m!) ways to order the beads.

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Necklaces are usually considered equivalent up to rotation, so this answer overcounts. – sdcvvc Oct 14 '12 at 16:55
Sorry, I was not aware of that! Thanks for correcting. – Jen-Ya Oct 14 '12 at 17:44

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