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Meg finishes ahead of Bob

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by sanju09 » Sat Feb 21, 2009 5:34 am
Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob?

A. 24
B. 30
C. 60
D. 90
E. 120
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by sureshbala » Sat Feb 21, 2009 5:45 am
The total number of different ways in which the race can be finished is 5! = 120.

The chance that Meg finishes ahead of Bob is 1/2.

Hence there will be 60 ways in which M finishes ahead of B
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by x2suresh » Sat Feb 21, 2009 6:00 am
sureshbala wrote:The total number of different ways in which the race can be finished is 5! = 120.

The chance that Meg finishes ahead of Bob is 1/2.

Hence there will be 60 ways in which M finishes ahead of B
agree with this approach
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by navami » Mon Sep 26, 2011 2:18 am
60
This time no looking back!!!
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