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Math Help - help me plz

  1. #1
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    help me plz

    i have never taken any probability classes but when i was studying for the SAT and saw this problem!!!!!!!! plz help me!!
    A committe is to consist of four members. if there are seven men and seven women avaiable to serve the committee, how many different committees can be formed?
    when i checked for the answer it was 1001, but i have no idea why?!!
    plz try to used simple math terms while solving the problem
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  2. #2
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    Quote Originally Posted by mooon91 View Post
    i have never taken any probability classes but when i was studying for the SAT and saw this problem!!!!!!!! plz help me!!
    A committe is to consist of four members. if there are seven men and seven women avaiable to serve the committee, how many different committees can be formed?
    when i checked for the answer it was 1001, but i have no idea why?!!
    plz try to used simple math terms while solving the problem
    You are choosing 4 people from 14 people .... 14C4.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    You are choosing 4 people from 14 people .... 14C4.
    thanks but i really don't know what that C means or how the answer is 1001. i tried to slove it, but i can't get it right. could you explain it more please.
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  4. #4
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    C means combinations. nCk =  \frac{n!}{k!(n-k)!}
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  5. #5
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    Quote Originally Posted by nzmathman View Post
    C means combinations. nCk =  \frac{n!}{k!(n-k)!}
    thanks that was really helpful
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  6. #6
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    Quote Originally Posted by mooon91 View Post
    thanks but i really don't know what that C means or how the answer is 1001. i tried to slove it, but i can't get it right. could you explain it more please.
    Do you have any experience with problems where you have to choose r objects from n objects ....? Read this: Combinatorics - GMAT Study Guide
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  7. #7
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    There are 14 people you may choose from so there are 14 ways to choose the first person on the commitee. That leaves 13 ways to choose the next person, 12 ways to choose the third, and 11 ways to choose the fourth.

    That means there are 14(13)(12)(11) ways to choose them IN THAT ORDER. But if we had 4 specific people and wanted to know how many different order the could be picked in we could use the same argument: there are 4 ways to decide who is first, 3 ways to decide who is second, 2 ways to decide who is third and 1 person left to be last: 4(3)(2)(1).

    Because in a committee (as opposed to choosing, say, a president, vicepresident, secretary and treasurer) order is not important, to discount those different orders for choosing the same 4 people we must divide by the number of different orders:
    14(13)(12)(11)/4(3)(2)1= 1001.

    Since 4(3)(2)1 can be abreviated "4!" we could also write that number as
    14(13)(12)(11)(10)...(3)(2)1/[{4(3)(2)(1)}{10(9)...(3)(2)(1)}] where I have multiplied both numerator and denominator by 10(9)...(3)(2)(1) so I can write it as 14!/(4!)(10!)= _14C_4.
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