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

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Manager
Joined: 10 Oct 2005
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Meg and Bob are among the 5 participants in a cycling race.  [#permalink]

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10 Dec 2005, 16:22
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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
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Kudos [?]: 78 [0], given: 0

Director
Joined: 10 Oct 2005
Posts: 713

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Location: Madrid
Re: PS - Combinations and arrangements - Tough [#permalink]

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11 Dec 2005, 10:35
TOUGH GUY 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

Total orders is 5!=120
total possible orders when Meg is ahead of Bob is Total-Bob is ahead of Meg
1 When Bob is first there are 4!=24 because it doesn't matter where mary is! B __ ___ __ __

2 When Bob is second 3!=6
__B M__ __
3 When Bob is third 3!=6
__ __ B M __
4 When Bob is forth 3!=6
__ ___ __ B M

So the cases when Mary is ahead of Bob 120-24-6-6-6=78
Where am I wrong!
Any help!

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Intern
Joined: 05 Sep 2005
Posts: 3

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11 Dec 2005, 12:01
Yurik,
I think your 2 and 3 are wrong. You have left out some possiblities.

Is the OA 60?

I got 60 this way:

Let A, B, C, D, M represent the five particants, where M - Meg and B - Bob

case 1: M finishes first.
M _ _ _ _ . There are 4! = 24 because it doesnt matter where B finishes. We just need to pick 4 from 4. ie. 4P4

case 2: M finishes second. There are 3 * 3! = 18 possibilities.
Possibilities are A M _ _ _ leaving 3! if A finishes first and M second. 2 more such possiblities
if C first and D first.

case 3: M finishes third. There are 2 * 3! = 12 possibilities.
Possibilities are _ _ M B _ giving 3P3 or 3! and _ _ M _ B again resulting in 3!

case 4: M finishes fourth. There are 3! = 6 possiblities.
Possiblities are _ _ _ M B giving 3P3 or 3!

So, the cases when M finishes ahead of B is : 24 + 18 + 12 + 6 = 60.

TOUGH GUY, can you post the OA please.

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Intern
Joined: 11 Sep 2005
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Location: Framingham, MA, USA

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11 Dec 2005, 13:06
M B - - - = 1 * 1 * 3 * 2 * 1 = 6
M - B - - = 1 * 3 * 1 * 2 * 1 = 6
M - - B - = = 6
M - - - B = = 6

- M B - - = 3 * 1 * 1 * 2 * 1 = 6
- M - B - = 3 * 1 * 2 * 1 * 1 = 6
- M - - B = = 6

and so on
Therefore,
6*4 + 6*3 + 6*2 + 6*1 = 6(4+3+2+1) = 6 * 10 = 60

Answer = C = 60

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Manager
Joined: 15 Aug 2005
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Re: PS - Combinations and arrangements - Tough [#permalink]

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11 Dec 2005, 19:31
TOUGH GUY 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

I agree with 60

the number of possible orders of the 5 runners are 5! =120.
Now, half the time Meg would be ahead of Bob and the other half time the opp. is true.
So no. of orders where Meg finishes ahead of Bob= 120/2= 60

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Manager
Joined: 10 Oct 2005
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Location: Hollywood

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12 Dec 2005, 06:04
OA is 60

(4!) + (3!*3) + (3!*2) + 3! = 60
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Beat the tough questions...

Kudos [?]: 78 [0], given: 0

12 Dec 2005, 06:04
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# Meg and Bob are among the 5 participants in a cycling race.

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