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Digits probability

  1. Jan 29, 2014 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    how many 5 digit unique numbers can you make from the following list: [itex]1,2,2,3,3,4,5,6,6,6,0[/itex] and I assume zero cannot be first.

    3. The attempt at a solution
    since there are 11 numbers, typically i would say [itex]\frac{10!*10}{2!*2!*3!}[/itex] which would be the numbers over their repeats, but the problem is we have only 5 spots, not 11, and hence my problem.

    thanks for your help!
     
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  3. Jan 29, 2014 #2

    haruspex

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    I suggest breaking it into cases according to the repeat counts of digits. I.e. start with "full house", 3 of one (6s obviously) and 2 of another. For each pattern, count the number of ways of supplying the digits and multiply by the number of ways of sequencing them.
    To deal with the zero, could permit it anywhere, then subtract off the number of 4-digit numbers not using the zero.
     
  4. Jan 30, 2014 #3

    joshmccraney

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    so something like this (for now i'll ignore the zero): [tex]\underbrace{{2\choose1}\frac{5!}{2!3!}}_{\text{both full houses}}+\underbrace{{3\choose2}{3 \choose 1}\frac{5!}{2!2!}}_{\text{3 of the two pairs without a lone 6}}+\underbrace{{3\choose1}\frac{5!}{2!}}_{\text{three of the single pairs}}+\underbrace{{6\choose5}5!}_{\text{no pairs}}[/tex] is this correct (or at least close)? my choose notation is for selecting which of what to use and the factorials are for counting the arrangements once we have chosen what to use. again, ive disregarded zero for now, but i'm not sure this is complete even without zero. any help please!
     
  5. Jan 30, 2014 #4

    haruspex

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    The full house term (3+2) I agree with. For 2+2+1, you have a choice of 3 digit values for the pairs (3C2), but the single digit can then be any digit value not chosen: 5C1. Similarly in the 2+1+1+1 term.
     
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