Joanna has six beads that she wants to assemble into a bracelet. Two of the beads have the same color, and the other four all have the same color. How many different ways can Joanna assemble her bracelet? (Two bracelets are considered identical if one can be rotated and/or reflected to obtain the other.)

Guest Mar 10, 2023

#1**0 **

We can solve this problem using casework.

Case 1: The two same-colored beads are adjacent

There are 2 ways to choose the color of the same-colored beads. Once we choose the color, there are 2 ways to arrange the same-colored beads, and then 4! / 2! ways to arrange the other beads. Therefore, there are 2 * 2 * (4!/2!) = 48 ways to assemble the bracelet in this case.

Case 2: The two same-colored beads are not adjacent

There are 2 ways to choose the color of the same-colored beads. Once we choose the color, there are 2 ways to arrange the same-colored beads, and then 4! ways to arrange the other beads. However, we have overcounted the arrangements where the bracelet can be rotated and/or reflected to obtain another arrangement. There are 6 ways to rotate the bracelet, and then 2 ways to reflect it, for a total of 12 equivalent arrangements. Therefore, there are (2 * 2 * 4!) / 12 = 16 ways to assemble the bracelet in this case.

Adding up the two cases, we get a total of 48 + 16 = 64 ways to assemble the bracelet.

Therefore, there are 64 different ways for Joanna to assemble her bracelet, if two of the beads have the same color and the other four have the same color.

Justingavriel1233 Mar 10, 2023