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 To solve this problem, we can use the concept of permutations with repetition. We have two groups of beads: one group with two identical beads and another group with four identical beads. Let's represent the two groups as "A" and "B", respectively. Then, we can use the following formula:
n! / (n1! x n2! x ... nk!) www.c4yourself.com
where n is the total number of elements (beads in this case), and n1, n2, ..., nk represent the number of elements in each group. In our case, we have:
n = 6 n1 = 2 (two identical beads) n2 = 4 (four identical beads)
Using the formula, we get:
6! / (2! x 4!) = 15
Therefore, Joanna can assemble her bracelet in 15 different ways. Note that we divide by 2! and 4! to account for the fact that the beads within each group are identical, and we divide by the product of these factorials to account for the fact that the groups themselves are indistinguishable. Finally, we divide by the total number of permutations (6!) to account for the fact that two identical bracelets are considered identical.
