How many ways to make a bracelet with $n$ white balls and $m$ black balls?

Question

Okay so I have $n$ black ball and $m$ white balls. How many bracelets can I make using all the balls? At first I thought there are $(n+m-1)!$ if all the balls were different. So we can divide into groups of $n!\cdot m!$ that look the same if you permute the black balls between them and also for the white balls.

However if I had $6$ white balls and $2$ black this would yield $3.5$ so clearly this can't be correct. How can I count this?

Is there a generalized version form for $k$ colors?

See: http://math.stackexchange.com/questions/600/circular-permutations-with-indistinguishable-objects/825#825 — Unreasonable Sin, Mar 12 '14 at 22:54

score 1 · Answer 1 · answered Mar 12 '14 at 22:55

If it is a straight line, it is $\frac {(n+m)!}{n!m!}={n+m \choose n}$ because you choose the positions for the white balls and the black ones follow. When you make a bracelet, naively you would divide by $n+m$ for the rotations, giving exactly the formula you state: $\frac {(n+m-1)!}{n!m!}$ The problem comes from patterns that can be rotated to give the same necklace in more than one way. So for $n=4,m=2$ the ${6 \choose 2}=15$linear strings are

$BBBBWW,BBBWBW,BBWBBW,BWBBBW,WBBBBW,\\BBBWWB,BBWBWB,BWBBWB,WBBBWB,BBWWBB,\\BWBWBB,WBBWBB,BWWBBB,WBWBBB,WWBBBB$

Now when we make necklaces of them, we get $BBBBWW,BBBWBW,BBWBBW$, where the first two account for six of the fifteen, but the last only accounts for three.

Oh, I see, but how can we count them then? – Asinomás Mar 12 '14 at 22:56 — Asinomás, Mar 12 '14 at 22:56

score 1 · Accepted Answer · answered Mar 12 '14 at 23:05

1

The complete solution to this is given by Pólya's theory of counting. The result is quite complex, no simple formula results except for very special cases.

answered Mar 12 '14 at 23:05

vonbrand

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How many ways to make a bracelet with $n$ white balls and $m$ black balls?

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