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# My friend lost 2 charms off her 7-charm bracelet. For her birthday, I bought her a new charm to replace one of the lost ones. Unfortunately,

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My friend lost 2 charms off her 7-charm bracelet. For her birthday, I bought her a new charm to replace one of the lost ones. Unfortunately, I messed up and got her a duplicate of one of the charms she still has. How many distinguishable ways can she put her 6 charms on her bracelet? (Two of the charms are the same, rotations are indistinguishable, and turning the bracelet front-to-back is indistinguishable.)

Guest Mar 29, 2015

#2
+4664
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Hi Melody,

I was willing to answer that thinking it had to do with Permutations but I didn't want to risk getting it wrong.

MathsGod1  Mar 29, 2015
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#1
+92206
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I think it is     $$\frac{5!}{2!}=60$$     that should take care of the 2 the same and the rotations.

If they were set in a line it would be 6!/2 but you always take one off when they are set in a circle.

But since you also have to take care of the turning front-to-back you might need to divide this by two.

So the answer might be 30

Melody  Mar 29, 2015
#2
+4664
+13

Hi Melody,

I was willing to answer that thinking it had to do with Permutations but I didn't want to risk getting it wrong.

MathsGod1  Mar 29, 2015
#3
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Just saying, the correct answer should be 30. You do need to divide by 2.

Guest Mar 24, 2016
#4
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Just saying, the correct answer should be 30. You do need to divide by 2.

Guest Mar 24, 2016

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