AMC 12 Mock Exam 5 Problem 17

There are $\frac{6!}{2!\,2!\,2!}=90$ colorings of six labeled positions. The bracelet's $12$ symmetries (6 rotations and 6 reflections) act on these colorings, and the number of distinct bracelets is the average number of colorings fixed by a symmetry (Burnside's lemma, or “each bracelet is counted once for each symmetry that fixes it”).

• Identity: $90$.
• Rotations by $\pm60^\circ$: $0$ (all beads would be one color). Rotations by $\pm120^\circ$: $0$ (each color would appear a multiple of $3$ times).
• Rotation by $180^\circ$: opposite beads match, so the three opposite pairs get the three colors: $3!=6$.
• Three reflections through two opposite beads: the two fixed beads get one color and the two swapped pairs get the other two colors: $3\cdot2=6$ each.
• Three reflections through midpoints of opposite edges: the three swapped pairs get the three colors: $3!=6$ each.

The average is $\frac{90+6+3\cdot6+3\cdot6}{12}=\frac{132}{12}=\boxed{11}$.