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

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Joined: 09 Sep 2013
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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24 Jun 2016, 07:07
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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24 Aug 2016, 12:12
marcodonzelli 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?

A. 24
B. 30
C. 60
D. 90
E. 120

this is similar to the problem, "How many 5 letter words can we form with a 5 letter word in which 2 letters are repeated, ex. ALPHA"
and the answer is $$\frac{5!}{2!}$$ = 60.
Meg following Bob can be treated as one group or identical and it will be exactly $$\frac{1}{2}$$ times of Total and equal to Bob following Meg.

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

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30 Aug 2017, 06:23
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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

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05 Sep 2017, 18:06
marcodonzelli 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?

A. 24
B. 30
C. 60
D. 90
E. 120

We can create the following equation:

total number of ways to finish the race = number of ways Meg finishes ahead of Bob + number of ways Meg does not finish ahead of Bob

Since the total number of ways to complete the race is 5! = 120, and since there are an equal number of ways for Meg to finish ahead Bob as there are for her not to finish ahead of Bob, Meg can finish ahead of Bob in 60 ways.

Alternate Solution:

If Meg comes in the first, then Bob can finish the race in any of the remaining positions, so there are 4! = 24 ways this could happen.

If Meg comes in the second, there are 3 choices for the first spot (since Bob can’t finish ahead of Meg) and the last 3 spots can be filled in 3! ways; so there are 3 x 3! = 3 x 6 = 18 ways this could happen.

If Meg comes in the third, there are 3 choices for the first spot, 2 choices for the second spot, 2 choices for the fourth spot, and 1 choice for the last spot; so, this could happen in 3 x 2 x 2 = 12 ways.

Finally, if Meg comes in the fourth, then Bob must finish last and the first 3 spots can be filled in 3! = 6 ways.

Note that Meg cannot finish the race in the last position because then Bob will have finished the race ahead of Meg.

In total, there are 24 + 18 + 12 + 6 = 60 ways Meg can finish the race ahead of Bob.

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Re: Meg and Bob are among the 5 participants in a cycling race.   [#permalink] 05 Sep 2017, 18:06

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