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Math Help - Combination/Permutation

  1. #1
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    Combination/Permutation

    A panel of judges is to consist of fourteen women and five men. A list of potential judges includes sixteen women and seven men. How many different panels could be created from this list?
    (A) 360
    (b) 5040
    (c) 2520
    (d) 158

    I am pretty sure that you use combination, but i can seem to get any of these numbers....could anyone please help me with this problem?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    A panel of judges is to consist of fourteen women and five men. A list of potential judges includes sixteen women and seven men. How many different panels could be created from this list?
    (A) 360
    (b) 5040
    (c) 2520
    (d) 158

    I am pretty sure that you use combination, but i can seem to get any of these numbers....could anyone please help me with this problem?
    Yes! you indeed have to use combinations. out of the 16 women, you must choose 14. out of the 7 men, you must choose 5. so the answer is "16 choose 14 * 7 choose 5"...or in math language: {16 \choose 14}\cdot{7 \choose 5}

    work that out and out have your answer
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  3. #3
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    So would it be 16 C 14?
    and 7 C 5?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    So would it be 16 C 14?
    and 7 C 5?
    Yes. _nC_r = C(n,r) = {n \choose r} they are all different notations for the same thing
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    I get 120 and 21?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by aikenfan View Post
    I get 120 and 21?
    you multiply them. that's what * means
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  7. #7
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    Thank you very much
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