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by prachi18oct » Sun Jul 12, 2015 10:18 am
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

Shouldn't the OA be D ? There should be exactly half of the total ways in which X is ahead of Y.
Please explain.
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by [email protected] » Sun Jul 12, 2015 11:40 am
Hi prachi18oct,

Assuming that there are no "ties", the 5 runners would have 5! = 120 different outcomes. Since Runner X will finish ahead of Runner Y half the time, the number of options is 120/2 = 60

Final Answer: D

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by theCEO » Sun Jul 12, 2015 1:07 pm
Alternative approach:

If X is in the 1st position - possibilities are 1 x 4 x 3 x 2 x 1 = 24
If X is in the 2nd position - possibilities are 3 x 1 x 3 x 2 x 1 = 18
If X is in the 3rd position - possibilities are 3 x 2 x 1 x 2 x 1 = 12
If X is in the 4st position - possibilities are 3 x 2 x 1 x 1 x 1 = 06
Total possibilities = 24+18+12+06=60
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