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Combination/Probability

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MBALA2009 Just gettin' started! Default Avatar
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Combination/Probability Post Thu Jul 10, 2008 9:49 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students, what is the probability that the number of men is equal to the number of women?

    A) 18/35
    B) 16/35
    C) 14/35
    D) 13/35
    E) 12/35

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    lightbulb Rising GMAT Star Default Avatar
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    Post Thu Jul 10, 2008 10:00 pm
    4M, 4W

    Need: 2M, 2W

    Favorable ways: C(4,2) * C(4,2) = 6*6 = 36
    Total ways: C(8,4) = 70

    Pr = 36/70 = 18/35

    Ans: A

    sudhir3127 GMAT Destroyer! Default Avatar
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    Post Fri Jul 11, 2008 4:01 am
    IMO answer is A,

    probability of choosing 2 boys out of 4 is 4C2 'lly girls 4C2.

    probability of chossing 4 students out of 8 is 8C4.

    hence its.. (4C2 x 4C2)/8C4 = 36/70= 18/35.

    Sudhir

    evansbd Rising GMAT Star Default Avatar
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    Post Fri Jul 11, 2008 11:28 am
    Can anyone suggest a way to do 8C4 by hand quickly?

    4C2 is simple enough if you remember factorials well....but the others, not so much.


    1! - 1
    2! - 2
    3! - 6
    4! - 24
    5! - 120
    6! - 720
    7! - 5040
    8! - 40320
    9! - 362880
    10! - 3628800

    BlueRain Rising GMAT Star Default Avatar
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    Post Sun Jul 13, 2008 10:42 pm
    8C4 = 8! / [4! x (8-4)!]
    =8! / 4! / 4!
    =8x7x6x5x4x3x2x1 / 4x3x2x1 / 4! (cancel 4x3x2x1 terms in 8!)
    =8x7x6x5 / 4x3x2x1 (cancel 4x2 with 8, divide 6 by 3 leaves 2)
    =7x2x5
    =70

    Most of the terms in permutation/combinations cancel due to sharing same tail terms (1x2x3xetc.), so you don't have to do the whole 8!.

    Hopefully this helps.

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