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.)
+11 Hello gm dear,
very nice question , well now we will see how it can be solved
There are two cases to consider: one where the two same-colored beads are next to each other, and one where they are not.
Case 1: Two same-colored beads are next to each other
In this case, we can treat the two same-colored beads as a single "unit". We then have five units to arrange: the "unit" of the two same-colored beads and the four units of the other color. The number of ways to arrange five units is 5! = 120. However, we must divide by 2! to correct for overcounting the arrangements of the two same-colored beads within the "unit". Therefore, there are 120/2! = 60 arrangements in this case.
Case 2: Two same-colored beads are not next to each other
In this case, we can treat the two same-colored beads as two separate units. We then have six units to arrange: the two units of the same-colored beads and the four units of the other color. The number of ways to arrange six units is 6! = 720. However, we must divide by 2! to correct for overcounting the arrangements of the two same-colored beads within their units, and we must also divide by 2 to correct for overcounting the arrangements that are reflections of each other. Therefore, there are 720/(2! * 2) = 180 arrangements in this case.
Overall, the total number of different ways that Joanna can assemble her bracelet is the sum of the two cases, which is 60 + 180 = 240.
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