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Math Help - probability/combinations & binomial theory

  1. #1
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    Exclamation probability/combinations & binomial theory

    there are 9 men, 12 women in the math club.

    how many possible committees of four people can be formed if Jill (woman) and Jack (man) cannot be on the same committee?

    how many four person committees have more women than men if Jill and Jack cannot be on the same committee?
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  2. #2
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    Hello, mathh18!

    There are 9 men, 12 women in the math club.

    How many possible committees of four people can be formed
    if Jill (woman) and Jack (man) cannot be on the same committee?
    With no constraints, there are: . {21\choose4} \,=\,5985 possible committees.

    How many committtees have both Jack and Jill?
    Since Jack and Jill are on the committee, choose 2 more from the other 19 people.
    . . There are: . {19\choose2} \,=\,171 committees with both Jack and Jill.

    Therefore, there are: . 5985 - 171 \:=\:\boxed{5814} committees
    . . in which Jack and Jill do not serve together.




    How many four-person committees have more women than men
    if Jill and Jack cannot be on the same committee?
    There are two cases to consider:
    . . (1) 4 women
    . . (2) 3 women, 1 man


    (1) 4 women
    . . There are: . {12\choose4} \:=\:{\color{blue}495} all-women committees.


    (2) 3 women, 1 man

    There are two cases to consider:
    . . (a) Jill is on the committee.
    . . (b) Jill is not on the committee.


    (a) Jill is on the committee.
    Choose 2 more women from the other 11 women: . {11\choose2} \,=\,55 ways.
    Choose one man from the 8 men (other than Jack): . {8\choose1} \,=\,8 ways.

    Hence, there are: . 55\cdot8 \:=\:{\color{blue}440} committees with 3 women, 1 man (with Jill).


    (b) Jill is not on the committee.
    Choose 3 women from the other 11 women: . {11\choose3} \:=\:165 ways.
    Choose 1 man from the available 9 men: . {9\choose1}\:=\:9 ways.

    Hence, there are: . 165\cdot9 \:=\:{\color{blue}1485} committees with 3 women, 1 man (without Jill).


    Therefore, there are: . 495 + 440 + 1485 \:=\:\boxed{2420} committees
    . . which have more women than men, and Jack and Jill do not serve together.

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