"bracelet type" Combinatorics

Question

This question seems ok but I'm having real difficultly working out the answer using the method they provided. It's so hard to keep track of all the options. Does anyone know of a better more algebraic method?

Answer 1

Two things can be done to a bracelet without changing it: rotating it and flipping it upside down. We want to calculate the number of possible bracelets without counting two bracelets that are identical up to rotation and flipping twice.

First, it is always possible to rotate the bracelet so that the yellow bead will be on top. It is left to determine the color of the 6 other beads, out of which 2 are red and four green.

Let's name the beads (clockwise) by 1-2-3-4-5-6. Note that 1-6, 2-5 and 3-4 are the symmetric bids (so when flipping the bracelet, the bids in 1-6 change places, 2-5 and 3-4).

In total, there are ${6 \choose 2}$ options to choose the red beads. Out of which, in 3 options, the red beads are in symmetric places.

In the rest ${6 \choose 2}-3$ options, each bracelet is counted twice due to this symmetry. Hence, divide by $2$ and the final answer is:

$$3+\tfrac{{6 \choose 2}-3}{2}=9$$

We can also list the options. If there is a red bead in place #1, the other red can be in any of the places, so there are 5 options. If there is a red bead in place #2, the other red can be only in places 3-5, because if it was in 6, it would be symmetric to the previous case with a bid at 1 and 5. So in total there are 3 options here. If there is a red bead in place #3, the only other place for the other red is #4, because other places were already taken into account in previous options.