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Numbers formed by taking each of the digits 2,3,5,6 and 8 once?

  1. Nov 5, 2003 #1
    What is the sum of all the numbers formed by taking each of the digits 2,3,5,6 and 8 once?
     
  2. jcsd
  3. Nov 5, 2003 #2
    haha, maybe later.
     
  4. Nov 6, 2003 #3

    jamesrc

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    6399936
     
  5. Nov 6, 2003 #4
    How'd ya get that answer? :smile:
     
  6. Nov 6, 2003 #5
    Hmm...combinatorics.

    2 3 5 6 8

    There are: (5)*(5-1)*(5-2)*(5-3)*(5-4) or 5*4*3*2*1 = 120 permutations for the above set of numbers.

    One set of numbers add up to: 2+3+5+6+8 = 24.

    So, the sum of all numbers formed from 120 permutations is:

    120*24 = 2880.
     
  7. Nov 6, 2003 #6

    jamesrc

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    If I answered the right question, for all of the 5-digit numbers you can make, each digit appears 24 times in each column, so you can write:

    24*(22222+33333+55555+66666+88888)=6399936

    Edit: Oh, I thought it was the sum of the numbers made by the combinations, not the sum of their digits.
     
    Last edited: Nov 6, 2003
  8. Nov 6, 2003 #7
    James gets the point!

    Here is an alternative way:
    4!*(2+3+5+6+8)*(11111) = 6,399,936
     
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