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

 Mar 20, 2020
edited by Guest  Mar 21, 2020
 #1
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A guest answered this here : https://web2.0calc.com/questions/help_44729, but it's incorrect

 Mar 21, 2020
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Thanks for pointing this old post out guest. :)

I have linked it to this one. 

Melody  Mar 21, 2020
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Start with a "reference bead" that we can use to indicate the position of each bead, with the reference bead being one of the distinct colored beads. Next, realize that of the 5 remaining beads left, the possible ways to select the same-colored beads is \({5 \choose 2}\)

= 10 ways to pick the pair. After you pick where the pair goes, the remaining beads have 4! possibilities to go to(do you see why? The first bead has 4 places, the second has 3, etc.) The total number of possibilities is then 10 * 24 = 240 arrangements. If you see something wrong with my solution, plz lmk. I haven't been the best historically with counting  ;).

 Mar 21, 2020
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Hi jfan17,

 

I am not brilliant or infallible on these counting questions either but I cannot follow your logic.

 

This is what I think.

Choose one of the beads to be a marker. It does not matter which one but I think the logic is easier to follow if you chose one of the individual ones. I will call it bead X

Now there are 5 beads left to place and one is a repeat so I think that will be  5!/2! = 60 permutations.

 

So far I have accounted for rotations being the same but I have not accounted for reflections.

The X bead is fixed.  There is only one axis of symmetry that allows this bead to keep the same position.

One reflection line makes two halves so I must divide by 2.

 

So I think that the answer is 30 possibilities.

 Mar 21, 2020
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Does anyone have the ability to check these answers?

 

I think it can be done in Wolfram|Alpha but I do not know how to phrase the question to get the answer......

 Mar 21, 2020
edited by Melody  Mar 21, 2020
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Thank you so much! Melody, your logic is correct!

 

This is what AOPS says : 

Solution:

The quick way to tackle this problem is correcting for overcounting. There are  6!/6 = 5! ways to arrange 6 beads in a circle. Since we can flip the bracelet, we have 5!/2 = 60 arrangements. Dividing by 2 once more to account for the fact that 2 beads are the same, we have 60/2 = 30 final arrangements.

We can also break the problem into three cases: the identical beads are adjacent, there's one bead between the identical beads, and there's two beads between the identical ones (i.e. they're opposite each other).

If the identical beads are adjacent, there are 4!/2=12 ways to order the other beads, where we divide by 2 because the bracelet can be flipped over to give another one of the initial 4! orderings (so these two orderings are indistinguishable)---hence, our 4! counts each possible arrangement twice. Make sure you see why we aren't dividing by anything for rotation---we don't count rotations of the same arrangement multiple times in our initial count of 4!.

If the two beads are one apart, there are 4 ways to choose the bead that is between them, and  ways to order the other three. As before, we can flip the bracelet to give another one of the initial  orderings, so there are 4!/2=12 ways to order the beads this way.

Finally, we have the two beads opposite each other. As before, we have 4! initial orderings. However, this time, there are 2 symmetries---the bracelet can be flipped and it can be rotated 180 degrees (thus swapping the positions of the two identical beads). Hence, each distinguishable arrangement is counted 4 times, so there are 4!/4=6 orderings.

Hence, there are 12+12+6=30 distinguishable ways for her to put the beads on her bracelet. If you still don't quite follow, break out a keychain and six keys and try it yourself.

 Mar 21, 2020
edited by MagnusMathematician  Mar 21, 2020

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