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by dunkin77 » Sun Jun 17, 2007 8:50 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


can anyone help??
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by jayhawk2001 » Sun Jun 17, 2007 1:44 pm
Is it 60 ?

I believe this was discussed in a recent thread. Total number of
combinations = 5! = 120. Out of this, half the number of times, M will
be before B and half the number of times B will be before M.

So, total = 120 / 2 = 60.

Alternatively, you can look at positions to compute this

When M is in first place, B can finish in 4 different places (i.e. MB---, M-B--
etc.). For each such case, we have 3! permutations for the other set of 3
people to finish the race.

So, when M is in first place, we have 4 * 3! possibilities.

Similarly, computing for M in second place, third place etc. we get

3! * (4 + 3 + 2 + 1) = 6 * 10 = 60 possibilities
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by dunkin77 » Sun Jun 17, 2007 3:34 pm
Thank you Jay. Yes, the answer is 60.
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by simplythebest » Mon Jun 18, 2007 2:25 am
Jay's method is the best...Still

Select 2 position from the total positions = 5C2

Assume that in the above position always Meg is ahead of the other

and the rest position can be filled in 3*2 ways

so total no of ways in which Meg finished before Bob

5C2 * 3* 2

and hence = 60
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