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Combinatorics-next problem with numbers

  1. Sep 15, 2005 #1
    Hi. It's one more hard task from cominatorics
    We have 5 digit. How many 7-digit numbers can we create that each two of them have at least 2 different digit?
    Could you help me?
    I think that the answer is 5^6 but don't know how to prove it.
     
  2. jcsd
  3. Sep 16, 2005 #2
    Could anybody help me? Please, give me at least hint.
     
  4. Sep 16, 2005 #3

    Tom Mattson

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    Let's start with this: Why do you think the answer is 56?
     
  5. Sep 16, 2005 #4
    OK.
    I don't know if you understand the task. The numbers have to have at least 2 different digit on some position. For example when we have 5 digit: 1, 2, 3, 4, 5, numbers 1234512 and 1234545 or numbers 5555555 and 1551555 are good.

    When we have 2-digit numbers we have 5^1=16 numbers and 5 that each two of them have at least 2 different digit.
    When we have 3-digit numbers we have 5^3=125 numbers and 5^2=25 that each two of them have at least 2 different digit because each two of 2-digit numbers have at least 1 different digit and when we add third number we get 25.
    So for 7-digit numbers we have 5^6.
     
    Last edited: Sep 16, 2005
  6. Sep 16, 2005 #5
    HOw is this different from yiour last post? and your english is a bit off.

    "can we create that each two of them have at least 2 different digit?" ??!??!
     
  7. Sep 16, 2005 #6
    Yes. you're right. My english isn't very good. But it should be clear now. Could anyone help me?
     
  8. Sep 17, 2005 #7
    Is there something that you can't understand in the problem or you don't know how to do it?
     
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