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Math Help - permutation question

  1. #1
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    permutation question

    In how many ways can 5 beads with different colours be sticked together by a circular wire?

    I thought in something round I must do like (5-1)!=4!=24, but the answer is 12..
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  2. #2
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    Re: permutation question

    I have always had trouble with circle permutations myself. Normally we would simply use (n-1)!. However, since this is a circular wire that can move around in space, you over count the permutations by 2. For problems like this, the required formula is (n-1)!/2. If you want an even better understanding, I would suggest researching the origins of the formula. With this formula, you can divide your 24 by 2 and get 12.
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  3. #3
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    Re: permutation question

    You can think it through like this:

    If the question was about 5 beads on a straight wire the answer qwould be 5!. Now when you join the neds of the wire into a loop you divide by 5 to eliminate the duplicates that you get by simply rotating the loop: for example the arrangment A-B-C-D-E is the same as B-C-D-E-A by simply rotating all the beads one position clockwise. I suspect that's how you got to 4! as your answer. But there is one more trick - for any arrangement you get an alternate mirror-image arrangement by turning the loop over. Thus A-B-C-D-E is the same as E-D-C-B-A., and so you have to divide the answer of 4! by 2, giving 12.
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    Re: permutation question

    Hello, Trefoil2727!

    Your answer is almost correct.
    But this problem has a twist that you haven't recognized.


    In how many ways can 5 beads with different colours be placed on a circular wire?

    I thought in something round I must do like: .(5 - 1)! = 4! = 24, but the answer is 12.

    ebaines is absolutely correct.

    If we are placing 5 distinct beads in a circle (on a table),
    . . the answer is 4! = 24.

    But we are placing the beads on a circular wire.
    . . That is, we are creating "bracelets".

    \text{So that: }\;\;\begin{array}{ccccc} &&A \\ \\ E &&*&&B \\ \\ &D&&C \end{array}\;\;\text{ is the same as }\;\;\begin{array}{ccccc}&&A \\ \\  B &&*&& E \\ \\ &C &&D \end{array}


    To eliminate duplication by reflection, we divide by 2.

    . . There are:. \frac{24}{2} \,=\,12 possible bracelets.
    Thanks from HallsofIvy and Trefoil2727
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