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GmatKiss GMAT Titan Default Avatar
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family Post Sun May 06, 2012 12:40 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?

    28
    32
    48
    60
    120

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    Bill@VeritasPrep GMAT Instructor
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    Post Sun May 06, 2012 1:49 pm
    The easiest way to solve this one is to find the total number of arrangements and then subtract the arrangements that violate the constraint.

    If the mother drives, there are 4 remaining people for 4 seats. 4!=24. We can swap the mother and father in each of these; 24*2=48 total arrangements.

    How do we violate the constraint? Have the two daughters sit next to each other. This must happen in the back seat, and it can occur in two ways:

    D-D-X

    X-D-D

    However, each daughter is unique, so we actually have:

    D1-D2-X
    D2-D1-X
    X-D1-D2
    X-D2-D1

    Let's consider the first one: D1-D2-X. We have two options for the driver's seat (M/F), two options for the other front seat(son and parent that's not driving), and one option for the other back seat, for a total of 4 arrangements.

    Each of the backseat options we identified has 4 arrangements, so 4*4=16 invalid arrangements.

    48-16=32 valid arrangements.

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    Anurag@Gurome GMAT Instructor
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    Post Sun May 06, 2012 7:21 pm
    GmatKiss wrote:
    A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there?

    28
    32
    48
    60
    120
    Daughters can be seated separately in two ways:

    (1) On the front seat in 2 ways. Others (including second daughter) can be arranged in 2 * 3! = 12 ways.
    Total no. of ways here = 2 * 12 = 24

    (2) Both are seated on the back seat by the window side.
    Others can be arranged in 2 * 2 * 1 = 4 ways.
    Total no. of ways here = 2 * 4 = 8

    Therefore, total seating arrangements = 24 + 8 = 32

    The correct answer is B.

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    Post Mon May 07, 2012 6:10 am

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