# In how many ways can you make a necklace with $n$ black beads and $m$ white beads? [duplicate]

Possible Duplicate:
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, FabianDec 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

(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.