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

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Director
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Meg and Bob are among the 5 participants in a cycling race.  [#permalink] New post 05 Jun 2005, 13:11
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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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Re: PS: Race [#permalink] New post 05 Jun 2005, 13:34
Vithal 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


Meg can come either 1st, 2nd, 3rd or 4th
If Meg is 1st, Bob can be in any of 4 places, the rest of the three participants can be in 3! arrangements
if Meg is 2nd, Bob can be in any of 3 places, rest -> 3!
If Meg is 3rd, Bob can be in any of 2 places, rest -> 3!
If Meg is 4th, Bob can only be in 1 place, rest -> 3!

hence total number of ways:
4*3! + 3*3! + 2*3! + 1*3!
(4+3+2+1)*3!
10*3! = 10*3*2 = 60

answer C
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Re: PS: Race [#permalink] New post 05 Jun 2005, 13:37
cloudz9 wrote:
Vithal 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


Meg can come either 1st, 2nd, 3rd or 4th
If Meg is 1st, Bob can be in any of 4 places, the rest of the three participants can be in 3! arrangements
if Meg is 2nd, Bob can be in any of 3 places, rest -> 3!
If Meg is 3rd, Bob can be in any of 2 places, rest -> 3!
If Meg is 4th, Bob can only be in 1 place, rest -> 3!

hence total number of ways:
4*3! + 3*3! + 2*3! + 1*3!
(4+3+2+1)*3!
10*3! = 10*3*2 = 60

answer C


that is where I got confused - fixing Meg is fine, but when you say Bob can be in any of 3 places and rest permuted in 3! ways, aren't you losing permutations where Bob can be moved?
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Re: PS: Race [#permalink] New post 05 Jun 2005, 13:41
Vithal wrote:
cloudz9 wrote:
Vithal 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


Meg can come either 1st, 2nd, 3rd or 4th
If Meg is 1st, Bob can be in any of 4 places, the rest of the three participants can be in 3! arrangements
if Meg is 2nd, Bob can be in any of 3 places, rest -> 3!
If Meg is 3rd, Bob can be in any of 2 places, rest -> 3!
If Meg is 4th, Bob can only be in 1 place, rest -> 3!

hence total number of ways:
4*3! + 3*3! + 2*3! + 1*3!
(4+3+2+1)*3!
10*3! = 10*3*2 = 60

answer C


that is where I got confused - fixing Meg is fine, but when you say Bob can be in any of 3 places and rest permuted in 3! ways, aren't you losing permutations where Bob can be moved?


well the way i see it...ok take the possibility that Meg came 2nd right?
it means that Bob can either be in 3rd place, 4th place or 5th place...which means 3 possibilites...that is taking care of everything
either _ M B _ _ or _ M _ B _ or _ M _ _ B where the _s represent the other 3 contestants...and they cna be arranged in 3! ways...whats the confusion then?
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Re: PS: Race [#permalink] New post 05 Jun 2005, 13:48
Vithal 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


Two possibilities, Meg after Bob or Bob after Meg, both occur in equal numbers
so

5!/2 = 60 or C
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Re: PS: Race [#permalink] New post 05 Jun 2005, 13:50
sparky wrote:
Vithal 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


Two possibilities, Meg after Bob or Bob after Meg, both occur in equal numbers
so

5!/2 = 60 or C


cool! that was much simpler reasoning! :)
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 [#permalink] New post 05 Jun 2005, 13:54
:oops got it!
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Re: PS: Race [#permalink] New post 05 Jun 2005, 18:11
Vithal 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

MB123=6x4=24
1MB23=6x3=18
12MB3=6x2=12
123MB=6x1=6
sum of the total = 60.
Re: PS: Race   [#permalink] 05 Jun 2005, 18:11
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