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Math Help - Committee of at least 3 from 20 people

  1. #1
    Member oldguynewstudent's Avatar
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    Committee of at least 3 from 20 people

    Given 20 people, how many ways are there to form a committee containing at least three people?
    \sum_{k=3}^{20}\left(\frac{20}{k}\right) is one answer but a shorter calculation would be 2^{20}-\left({20\atop 1}\right)-\left({20\atop 2}\right).

    Does this look correct?
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  2. #2
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    Hello, oldguynewstudent!

    Given 20 people, how many ways are there to form
    a committee containing at least three people?

    \sum_{k=3}^{20}\left(\frac{20}{k}\right) is one answer but a shorter calculation would be 2^{20}-\left({20\atop 1}\right)-\left({20\atop 2}\right).

    Does this look correct? . . . . um, not quite

    I assume you mean: . \sum^{20}_{k=3}{20\choose k}

    Also, the alternate way is: . 2^{20} - {20\choose0} - {20\choose1} - {20\choose2}

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  3. #3
    Member oldguynewstudent's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, oldguynewstudent!


    I assume you mean: . \sum^{20}_{k=3}{20\choose k}

    Also, the alternate way is: . 2^{20} - {20\choose0} - {20\choose1} - {20\choose2}
    Thanks, yes because you can't have a committee of no people.
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