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Figuring out # of possible combos of 3

  1. Sep 7, 2004 #1

    Math Is Hard

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    A friend asked me this question, and I am not sure I know how to answer it:


    ************************************************
    I have 15 items, and I want to figure out how many unordered combinations of three there are. Any ideas?

    a b c d e f g h i j k l m n o

    Example:
    abc bcd cde
    abd bce cdf
    abe bcf cde
    abf bcg cdg

    ************************************************


    The way I think it can be done is using this formula:
    [tex]
    {n\choose k}={n!\over k!(n-k)!}
    [/tex]

    so you would have [tex]{15!\over 3!12!}[/tex] or 455 possible combinations.
    But I am not really confident in this because I never studied this formula in a class - and I don't know if I am using it correctly.

    Thanks!
    -MIH
     
  2. jcsd
  3. Sep 7, 2004 #2

    HallsofIvy

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    Think about how you would choose such combinations. Start by choosing the first letter: there are 15 ways to do that. Now choose the second letter: 14 ways. Finally, choose the third letter: 13 ways. There are 15*14*13 ways. You could write that as 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1/12*11*10*9*8*7*6*5*4*3*2*1= 15!/12!= 15!/(15-3)!

    But that counts "abc" and "bac" as separate which you don't want to do. For each 3 letters there are 3! ways to get those same letters in different orders. To account for that, divide the answer above by 3!. That gives [tex]{15!\over 3!12!}[/tex]
    exactly as you suggest. The same analysis for k objects chosen from n gives [tex]{n\choose k}[/tex] (which, in tex form is written "n choose k"!).
     
  4. Sep 7, 2004 #3

    Math Is Hard

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    Thanks so much, Halls!!!
    I appreciate you taking the time to show me how that works.
     
  5. Sep 14, 2004 #4
    Hey does anyone know physics???
     
  6. Sep 14, 2004 #5

    Gokul43201

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    Sharayah, if you have a physics question, post in under General Physics, and create a New Thread.
     
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