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Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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23 Oct 2006, 10:43

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34% (01:01) wrong based on 148 sessions

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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?

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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23 Oct 2006, 10:56

girikorat wrote:

If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

It should be 4! = 24. Basically since Jen and Bob's positions are fixed - we can just treat them as ONE combination. Taking the remaining three folks + ONE combination of Jen and Bob you have 4 permutations for the race to finish.

Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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12 Aug 2014, 01:41

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girikorat wrote:

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?

Total # of ways the race can be finished is 5!. In half of the cases Meg finishes ahead of Bob and in other half Bob finishes ahead of Meg. So, ways Meg to finish ahead of Bob is 5!/2=60.

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