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Combinations! How many.

  1. Jun 2, 2007 #1
    Combinations! How many.........

    1. The problem statement, all variables and given/known data
    How many letter combinations with 9 letters are you able to make with following letters : M-A-T-E-M-A-T-I-K?


    2. Relevant equations
    Well its pretty obvious you need to use Combinations.

    Please explain how you solve this problem, don't write use combinations :smile:.
    I need to know how you think and from which angle you "attack" the problem.
     
  2. jcsd
  3. Jun 2, 2007 #2

    Dick

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    You have 9 letter positions to fill. First lets place the M's. There are two of them, so I have C(9,2) ways. Now lets do the A's. There 2 of them and 7 places left to fill, so I have C(7,2) ways. So far I've got C(9,2)*C(7,2). Can you finish?
     
  4. Jun 3, 2007 #3
    So iit´s C(9,2)*C(7,2)*C(5,2)*3*2*1?

    In my math book the answer is C(9,2)*C(7,2)*5!..... which i find a little strange.
     
  5. Jun 3, 2007 #4

    danago

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    When i did it, i got the same answer as you, and then to check, i got mathematica to output every single permutation of those letters into a list. That list contained 45360 elements, so unless ive misunderstood the question, it seems that you may be right.
     
  6. Jun 3, 2007 #5

    Hurkyl

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    Did you notice those are the same thing?

    Incidentally, it seemed most clear to me to write the answer as
    9! / (2! * 2! * 2!),​
    or, as a multinomial coefficient,
    [tex]\binom{9}{2 \ 2 \ 2 \ 1 \ 1 \ 1} .[/tex]​
     
    Last edited: Jun 3, 2007
  7. Jun 3, 2007 #6

    Dick

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    They aren't the same thing. They differ by a factor of two. Whoever wrote the solution seems to have miscounted the number of doubled letters.
     
  8. Jun 3, 2007 #7

    Hurkyl

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    Ah, right. This is what was written:
    C(9,2)*C(7,2)*C(5,2)*3*2*1​
    and this is what I thought I read:
    C(9,2)*C(7,2)*C(5,2)*3!*2!*1!​
     
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