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Math Help - Permutations

  1. #1
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    Permutations

    I encountered this question in some past papers & was wondering if anyone could guide me with it:

    A committee of 4 is drawn from a shortlist of 8 women & 6 men. How many possible committees are there? How many possible committees are there with 2 men and 2 women?

    Any help would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by callie View Post
    I encountered this question in some past papers & was wondering if anyone could guide me with it:

    A committee of 4 is drawn from a shortlist of 8 women & 6 men. How many possible committees are there? How many possible committees are there with 2 men and 2 women?

    Any help would be greatly appreciated.
    1) 4 if i were to answer you question.
    but i think you mean how many different ways of chosing the committees.

    so it would be 12C4 = 495.

    2) same again 4 if i were to answer you question.

    but the no. of different ways is 8C2 x 6C2 = 420.
    Last edited by mr fantastic; December 14th 2009 at 03:58 AM.
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  3. #3
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    Quote Originally Posted by callie View Post
    I encountered this question in some past papers & was wondering if anyone could guide me with it:

    A committee of 4 is drawn from a shortlist of 8 women & 6 men. How many possible committees are there? How many possible committees are there with 2 men and 2 women?

    Any help would be greatly appreciated.
    You can choose any one of the women to be on the committee. There are, of course, 8 ways to do that. Now you can choose any one of the remaining 7 women to be the second woman on the commitee. There are 7 ways to do that and so 8*7 ways to make those choices. But that counts "Betty first then Sarah second" as different from ""Sarah first, then Betty" which you don't want to do. So divide 2 since for each pair of women there are two orders.

    Do the same thing for the two men.
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