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Math Help - Simple Arrangements and Selections - committee question

  1. #1
    DBA
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    Simple Arrangements and Selections - committee question

    How many ways can a committee be formed from 4 men and 6 woman with:

    c) 5 people, and not all of the 3 O'Hara sisters can be on the committee.
    The answer is C(10,5) - C(7,2)

    I do not understand where the C(7,2) comes from. Can someone explain this?


    I tried to break it up into committees of 5 people:
    0 men - 5 woman ---> I can only choose out of 5 women since one sister of 3 sisters is out
    1 men - 4 women --> choose out of 5 women
    2 men - 3 women --> choose out of 5 women
    3 men - 2 women--> choose out of 6 women, since all 3 sisters cannot be chosen anyway
    4 men - 1 women --> choose out of 6 women
    5 men - 0 women --> choose out of 6 women

    and in the end I add up.
    C(4,0)*C(5,5) + C(4,1)*C(5,4) + C(4,2)*C(5,3) + C(4,3)*C(6,2) + C(4,4)*C(6,1)

    But I get a different number, so this must be wrong.

    Thanks for any help
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  2. #2
    DBA
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    Actually, I can answer it myself. I just figured it out.
    The total amout to create a committee with 4 men and 6 women = 10 people is C(10,5)
    The forbitten committees have the 3 sisters. So 3 of the 5 places in the committee are occupied by the sisters and we choose only people for the last two places.
    We have 7 people to choose from (10 - 3 sisters = 7), so it is C(7,2)

    Then we just substract the forbitten committees from the total possible committees C(10,5) - C(7,2)
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  3. #3
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    Hello, DBA!

    Seeing your first try, I got curious . . .


    How many ways can a committee be formed from 4 men and 6 woman with:

    c) 5 people, and not all of the 3 O'Hara sisters can be on the committee?

    Suppose the three sisters are A, B and C.
    And the other three women are x, y and z.


    There are five cases to consider . . .


    (1) 4 men, 1 woman
    . . .Choose all 4 of the men: C(4,4) = 1 way.
    . . .Choose any 1 of the 6 women: C(6,1) = 6 ways.
    . . .Committees with 4 men and 1 woman: 1 x 6 = 6

    (2) 3 men, 2 women
    . . Choose 3 of the 4 men: C(4,3) = 4 ways.
    . . Choose any 2 of the 6 women: C(6,2) = 15 ways.
    . . Committees with 3 men and 2 women: 4 x 15 = 60

    (3) 2 men, 3 women
    . . Choose 2 of the 4 men: C(4,2) = 6 ways.
    . . Choose 3 of the 6 women:
    . . . . Choose xyz: C(3,3) = 1 way
    . . . . Choose 2 of xyz, 1 of ABC: C(3,2) x C(3,1) = 9 ways
    . . . . Choose 1 of xyz, 2 of ABC: C(3,1) x C(3,2) = 9 ways
    . . . . There are: 1 + 9 + 9 = 19 ways to choose 3 women.
    . . Committees with 2 men and 3 women: 6 x 19 = 114

    (4) 1 man, 4 women
    . . Choose 1 of the 4 men: C(4,1) = 4 ways.
    . . Choose 4 of the 6 women.
    . . . . Choose xyz, choose 1 of ABC: C(3,3) x C(3,1) = 3 ways
    . . . . Choose 2 of xyz, choose 2 of ABC: C(3,2) x C(3,2) = 9 ways
    . . . . There are: 3 + 9 = 12 ways to choose 4 women.
    . . Committees with 1 man and 4 women: 4 x 12 = 48

    (5) 5 women
    . . Choose 3 of xyz, choose 2 of ABC: C(3,3) x C(3,2) = 3 ways
    . . Committees with 5 women: 3


    Therefore, there are: .6 + 60 + 114 + 48 + 3 .= .231 committees.

    . . . ta-DAA!



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