8. Lieneke makes beaded bracelets. Each bracelet has six beads: two white, two gray, and two black beads. Some bracelets may seem different at first glance, but they are not: if you rotate or flip one, it turns out to be the same as the other. For example, the three bracelets below are the same.
![](https://cdn.mathpix.com/cropped/2024_04_17_cc2ba9e3b1603277d398g-2.jpg?height=174&width=882&top_left_y=2397&top_left_x=588)

How many truly different bracelets can Lieneke make?
A) 10
B) 11
C) 12
D) 14
E) 15

## B-questions
A8. B) 11 We divide the bracelets into three groups: (I) the black beads are next to each other, (II) there is one bead between the two black beads, and (III) the black beads are opposite each other. In each case, we count how many different ways we can place the remaining beads.
![](https://cdn.mathpix.com/cropped/2024_04_17_2ddd9533af815a0f4acbg-3.jpg?height=200&width=1014&top_left_y=420&top_left_x=527)

Group (I) We need to choose which two of the positions $a, b, c, d$ will be white beads. There are six possibilities: $a b, a c, a d, b c, b d$, and $c d$. The choice $b d$ gives the same bracelet as $a c$, and the choice $c d$ gives the same bracelet as $a b$. In total, we find 4 different bracelets in group (I).

Group (II) Suppose we choose $a$ to be white. This gives two different bracelets: $a c$ white and $a b$ white. If we choose $a d$ white, we get the same bracelet as when we choose $a b$ white. If we choose $a$ to be gray, we also get two bracelets. Group (II) thus has 4 different bracelets.

Group (III) We find 3 different bracelets: $a b$ white, $a c$ white, or $a d$ white. The other three possibilities $(b c, b d, c d)$ give the same bracelet as (respectively) $a d, a c$, and $a b$.

In total, there are $4+4+3=11$ different bracelets.