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Math Help - A Permutation Question involving B

  1. #1
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    A Permutation Question involving B

    The Problem: How many bracelets can be made by putting six different-coloured beads on a ring if 10 different colours are available?

    Attempted Solution: Since there are 10 possible options and you must arrange six of the colours at a time to form a bracelet, I thought I should just use the permutation formula.

    10 subscript P subscript 6 = 151 200

    Since I bracelet may be fliped over, I then divided the answer by 2 to obtain 75 600.

    But the correct answer is 12 600.

    I am perplexed by how this could be. Could someone please help?

    Note: Sorry for puting "subscript". I'm not a master of this forum's codes.
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  2. #2
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    Hello, D. Martin!

    How many bracelets can be made by putting six different-coloured beads
    on a ring if 10 different colours are available?

    Attempted Solution: Since there are 10 possible colours and you must arrange six of them,
    I thought I should just use the permutation formula: . _{10}P_6 \:=\:151,\!200

    Since the bracelet may be flipped over, I then divided the answer by 2 to obtain 75,600.
    Good! . . . Most of us forget that.

    But the correct answer is 12,600.

    Since the beads are in a circle, an arrangement of 6 beads can be rotated
    . . without creating another bracelet.



    . . \begin{array}{ccccccc} & & A & - & B \\ & / & & & & \backslash \\ F & & & & & & C \\ & \backslash & & & & / \\ & & E & - & D \end{array}. . . . \begin{array}{ccccccc} & & F & - & A \\ & / & & & & \backslash \\ E & & & & & & B \\ & \backslash & & & & / \\ & & D & - & C \end{array}. . . . \begin{array}{ccccccc} & & E & - & F \\ & / & & & & \backslash  \\ D & & & & & & A \\ & \backslash & & & & / \\ & & C & - & B \end{array}



    . . \begin{array}{ccccccc} & & D & - & E \\ & / & & & & \backslash \\ C & & & & & & F \\ & \backslash & & & & / \\ & & B & - & A \end{array} . . . . \begin{array}{ccccccc}& & C & - & D \\ & / & & & & \backslash  \\ B & & & & & & E \\ & \backslash & & & & / \\ & & A & - & F \end{array} . . . . \begin{array}{ccccccc}& & B & - & C \\ & / & & & & \backslash \\ A & & & & & & D \\ & \backslash & & & & / \\ & & F & - & E \end{array}



    These six bracelets are identical . . . So your answer is too large by a factor of 6.

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