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Math Help - combination problem

  1. #1
    Junior Member
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    combination problem

    Have the following problem to solve, is this the correct method,

    how many ways can a subcommittee of 5 be selected from a committee of 12,

    so,,

    12C5 = 12!/7!.5!

    =479001600/(5040)x(120)

    is this right?
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  2. #2
    Senior Member
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    Hi

    Quote Originally Posted by offahengaway and chips View Post

    12C5 = 792

    is this right?
    Yes, it is
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  3. #3
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    so if i had to see how many 11 letter combinations out of 17 letters could be made would that look like this

    17C11 = 17!/6!.11!
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  4. #4
    Senior Member
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    Quote Originally Posted by offahengaway and chips View Post
    so if i had to see how many 11 letter combinations out of 17 letters could be made would that look like this

    17C11 = 17!/6!.11!
    Yes & No.

    Yes if you talk about 17 different letters

    There ain't 17C11 combinations if you have 12x X and 5x Y

    So it depends on your letters you are talking about.
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