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Permutation and combination

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    how many 4 digit numbers are there which do not contain more than 2 different numbers?


    2. Relevant equations




    3. The attempt at a solution
    all can contain the same digit
    any 3 out of the 4 places can be occupied by the same digit
    and
    any 2 out of the 4 places can be occupied by the same digit
     
  2. jcsd
  3. Oct 8, 2012 #2

    You are on the right track, think about it this way. You have a choice of choosing the numbers from 1 to 9 and then you have three cases:

    1. the case where all 4 numbers are the same number as you chose

    2. the case where 3 numbers are the same number as you chose as well as you have to choose another number that's different from the other three

    3. the case where 2 numbers are the same number as you chose and you have to choose 2 other numbers where they are different from each other and different from the other 2 you previously chose

    Does that make sense?
     
  4. Oct 8, 2012 #3

    rcgldr

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    Shouldn't that be digits from 0 to 9 (a total of 10 possible digits)?
     
  5. Oct 8, 2012 #4

    Ray Vickson

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    Probably yes, but (presumably) the first (left-most) digit cannot be 0. (Since this is not made clear, if I were doing the problem I would solve both versions.)

    RGV
     
  6. Oct 8, 2012 #5

    rcgldr

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    As you mention, it's not clear if leading zeroes are allowed, like a combination lock with 4 digits. I would assume they are allowed, since otherwise it eliminates all the combinations such as 0xxx, 00xx, 000x, 0000.
     
  7. Oct 8, 2012 #6
    Yes that was a typo, I did mean from 0 to 9 but I am not too sure if the first number would be allowed to be a 0. I would personally think otherwise, not letting the first number to 0
     
  8. Oct 8, 2012 #7
    as per the answer given, you will need to consider both the cases!
     
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