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Simple Probability Question

  1. Sep 2, 2012 #1
    Problem Statement:
    How many 5 digit number can be formed from the integers 1,2, ..., 9 if no digit can appear more than twice?

    Approach:
    18!/((2!*9)(18-5)!)

    Reason:
    Using the combination formula, we were able to approach this problem by saying that there are 18 numbers from which only 5 are chosen and the order does not matter for individual groups of two numbers.

    Is this approach correct? I would love a very detailed explanation of this problem since many of the ones in class are about this same difficulty level.
     
  2. jcsd
  3. Sep 2, 2012 #2

    haruspex

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    That answer (57120) is close, but doesn't allow properly for the different combinations of one or two pairs the same etc.
    In general, there's no easy way to solve these. For this one I would start with the full 95 and remove the disallowed:
    - three the same, two different: 9 * 8 * 7 * 5C2
    - "full house": 9 * 8 * 5C2
    - four the same: 9 * 8 * 5C1
    - five the same: 9
    Result: 52920
     
  4. Sep 4, 2012 #3
    Could you elaborate on what you mean by "one or two pairs the same, etc"?
     
  5. Sep 5, 2012 #4

    haruspex

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    You set up a model in which each digit is represented twice, as 1a, 1b, 2a, 2b etc. say.
    Your 18!/(18-5)! is the number of ways of choosing 5, in order, from 18. That's too many for two reasons: because choosing a single '1' will be counted twice (1a, 1b) and because choosing 1a for a certain position and 1b for another gives the same result as if they were swapped around. You can correct for the first problem by dividing by 32 (25, not 2!*9), but that overcorrects when both the a and b copies are used (divides by 2 for each of the two digit positions instead of just once for the pair).
    If we subdivide 18!/(18-5)! into three cases:
    X = number with no repeated digits
    Y = number with one digit repeated
    Z = number with two digits repeated
    then the final answer will be X/32 + Y/16 + Z/8
    So we can see the answer lies between (18!/(18-5)!)/8 and (18!/(18-5)!)/32. You happened to divide by 18, so got a number that wasn't too far off.
    Btw, X = 18 * 16 * 14 * 12 * 10 = 483840, Y = 9 * 5 * 4 * 16 * 14 * 12 = 483840, Z = 9C2 * 5 * 4 * 3 * 2 * 14 = 60480.
    X + Y + Z = 18!/(18-5)!; X/32 + Y/16 + Z/8 = 52920.
     
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