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GmatKiss GMAT Titan Default Avatar
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Runners Post Sat May 05, 2012 9:09 pm
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  • Lap #[LAPCOUNT] ([LAPTIME])
    Runners V, W, X, Y and Z are competing in the Bayville local triathlon. How many different ways are there for X to complete the race ahead of Y?
    (A) 5
    (B) 10
    (C) 30
    (D) 60
    (E) 120

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    sam2304 GMAT Titan
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    Post Sat May 05, 2012 9:28 pm
    X can take 1st/2nd/3rd and 4th position respectively. Other players have the option to come ahead of X but Y should always come behind X.

    X in first position with Y anywhere behind X
    X - - - - => 4!

    X in 2nd pos with any one of the three runners other than Y in 1st pos and the rest behind X
    3 X - - - => 3.1.3!

    X in 3rd pos with any one of the three runners other than Y in 1st pos and 2nd pos the rest behind X
    3 2 X - - => 6.1.2!

    X in 4th pos with Y behind X
    3 2 1 X 1 => 6

    24 + 18 + 12 + 6 = 60

    IMO D.

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    aneesh.kg GMAT Destroyer!
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    Post Sat May 05, 2012 11:29 pm
    Simplest method:

    These five people could have been arranged in 5! = 120 ways had there been no condition on them.
    The question imposes a condition that X has to be before Y.

    Come to think of it, in all the 120 arrangements X will be ahead of Y in 50% of the ways and Y will be ahead of X in 50% ways. There is no bias for any of them, it has to be half-half.

    So, X will be ahead of Y in (1/2)*(120) = 60 ways.

    (D) is the answer.

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