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Math Help - Ways of Selection Problem

  1. #1
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    Ways of Selection Problem

    Question: Security combination lock has 8 touch pads. Each lock is 3 digit code.

    How many different codes are possible if the order is unimportant, and the digits can be repeated?

    The answer here has 8 x 8 x 8.

    But I think this is wrong, as this orders security code. I.e. its counting repeats.

    For example, it would have counted 8 8 7 as different to 7 8 8 which will still work.

    Am I flawed in my reasoning? If not how do i remove the ordering.
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    Question: Security combination lock has 8 touch pads. Each lock is 3 digit code.
    How many different codes are possible if the order is unimportant, and the digits can be repeated?
    The answer here has 8 x 8 x 8.
    But I think this is wrong, as this orders security code. I.e. its counting repeats.
    If the question actually states that order is unimportant then that answer is not correct.

    The answer then is \displaystyle \binom{3+8-1}{3}.
    That is simply selecting three from eight choices.
    Last edited by Plato; August 9th 2010 at 08:00 AM.
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    But that dosent count repeition. I.e. the nos could be 888.
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  4. #4
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    It most certainly does count repetitions. It is the multi-selection formula.
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  5. #5
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    So sorry, Ive miss read your solution. I thought it was 8C3 (combinations) which was a bit strange.

    But I do not understand how you acquired the 10C8 whic does not equate to 8C3.
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    You want to read about multisets .
    Go down the page to the section on Multiset coefficients.
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  7. #7
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    Quote Originally Posted by Lukybear View Post
    Question: Security combination lock has 8 touch pads. Each lock is 3 digit code.

    How many different codes are possible if the order is unimportant, and the digits can be repeated?

    The answer here has 8 x 8 x 8.

    But I think this is wrong, as this orders security code. I.e. its counting repeats.

    For example, it would have counted 8 8 7 as different to 7 8 8 which will still work.

    Am I flawed in my reasoning? If not how do i remove the ordering.
    Suppose the pads are numbered 1 to 8.

    Why is the solution not

    \binom{8}{3}+8(8)\ ?

    Ah!!

    I get it!
    Last edited by Archie Meade; August 9th 2010 at 09:37 AM.
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