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

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Manager
Joined: 07 Jun 2006
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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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Difficulty:

25% (medium)

Question Stats:

65% (02:46) correct 35% (01:07) wrong based on 204 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?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Aug 2014, 01:40, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
Manager
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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23 Oct 2006, 10:46
I got 33 here

24+6+2+1, is it rite?
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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.
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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23 Oct 2006, 11:15
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Is it 60?
When Jen is in posion 1:
4! ways to arrange the other 4.

When Jim is in pos 2:
Bob has to be in 3rd, 4th or 5th pos. And the other 3 can be positioned in 3! ways
So 3x3!

When Jen is in pos 3:
Bob has to be in 4th ot 5th. Others can be positioned in 3! ways
so 2x3!

When Jen is in 4th:
Bob has to the 5th person. Others- 3!

So total = 4! + 3x3! + 2x3! + 3!
= 4! + 3!x6 = 24 + 36 = 60
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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23 Oct 2006, 11:20
yeah its 60....I think I got it now....

its 4*3!+3*3!+2*3!+1*3! = 60

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

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08 Aug 2014, 01:03
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Race can finish in 5!=120 ways, and exactly half of them Jen always finishes in front of Bob. The correct answer is 60.
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Last edited by smyarga on 08 Aug 2014, 13:31, edited 1 time in total.
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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08 Aug 2014, 12:54
smyarga wrote:
Race can finish in 6!=120 ways, and exactly half of them Jen always finishes in front of Bob. The correct answer is 60.

Think you mean 5! = 120
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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08 Aug 2014, 13:32
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sismail wrote:
smyarga wrote:
Race can finish in 6!=120 ways, and exactly half of them Jen always finishes in front of Bob. The correct answer is 60.

Think you mean 5! = 120

Yep:) edited) thanks!
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I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

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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?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html

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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OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
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Re: Meg and Bob are among the 5 participants in a cycling race.   [#permalink] 12 Aug 2014, 01:41
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