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Six identical white beads, three identical blue beads, and five identical red beads are to be strung together to create a bracelet. If the beads are free to move all the way around the bracelet, how many different bracelets can be made?

I was able to get $\frac{13!}{6! \cdot 5! \cdot 3! \cdot 2}$ = 6006, which I think accounts for the rotational and mirror symmetry, but this is apparently an undercount.

I was told to use Burnside's theorem, and that the answer is between 6006 and 7007. Any help would be really appreciated. Thanks!

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  • $\begingroup$ Um... There are 14 beads my friend, so it should be $14!$ on the top $\endgroup$
    – Aarony Jamesys
    Jul 2, 2020 at 3:07

2 Answers 2

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Since you were told to use Burnside, we look at the 28 different symmetries of the bracelet, and see how many bead arrangements are unchanged for each of them:

  • Rotating 1, 3 or 5 units in either direction. Any of these require all beads to be the same colour. Impossible.
  • Rotating 2, 4 or 6 units in either direction. Any of these require every other bead to be the same colour. Impossible.
  • Rotating 7 units. This requires the 14 beads to be partitioned into pairs where both beads in any pair has the same colour. Impossible.
  • Rotating 0 units. Here there are $\binom{14}{6}\binom{8}{3}\binom{5}{5} = 168\,168$ possibilities.
  • Mirroring along any of the 7 axes that go between the beads. Same as for the 7-unit rotation: we require 7 pairs of same-colour-beads. Impossible.
  • Mirroring along any of the 7 axes that intersect two beads. Here one of the two beads must be red, and the other blue. The remaining 12 beads must be divided up into 6 same-colour-pairs which are then placed along the bracelet. $2\cdot \binom{6}{3}\cdot\binom{3}{1}\cdot \binom22 = 120$ possibilities.

This means there are a total of $$ \frac{20\cdot 0 + 7\cdot120 + 168\,168}{28} = 6036 $$ distinct bracelets.

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Hint:

You should be able to convince yourself that none of the rotations apart from the identity could possibly when applied to an arrangement result in an identical arrangement.

Mirror symmetries where there are seven beads to either side of the axis of symmetry, i.e. the axis of symmetry passes through spaces between beads, also cannot result in an identical arrangement as there would necessarily be an odd number of blue beads on one side and an even on the other.

The mirror symmetry where beads lie along the axis of symmetry however can result in an identical arrangement.

Consider the following arrangement:

$$\begin{array}{cccccccc}~&\color{blue}{*}&*&*&*&\color{red}{*}&\color{red}{*}\\\color{blue}{*}&-&-&-&-&-&-&\color{red}{*}\\~&\color{blue}{*}&*&*&*&\color{red}{*}&\color{red}{*}\end{array}$$

(black is substituted in place of white)

The mirror symmetry when flipping along the horizontal axis will result in an identical arrangement. Similarly, there will be other arrangements which are symmetrical with regards to flipping. You should be able to convince yourself that each of these must have one of the beads along the axis of symmetry be blue and the other be red since the number of blue and red beads are both odd numbers.

I get an answer of $6036$ possible arrangements.

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