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by GmatKiss » Sat May 05, 2012 8:09 pm
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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by sam2304 » Sat May 05, 2012 8:28 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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by aneesh.kg » Sat May 05, 2012 10:29 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.

[spoiler](D)[/spoiler] is the answer.
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