It is currently 23 Jun 2017, 09:12

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
Director
Joined: 01 Feb 2003
Posts: 844
Meg and Bob are among the 5 participants in a cycling race.  [#permalink]

### Show Tags

05 Jun 2005, 14:11
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Senior Manager
Joined: 17 May 2005
Posts: 271
Location: Auckland, New Zealand

### Show Tags

05 Jun 2005, 14:34
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

Director
Joined: 01 Feb 2003
Posts: 844

### Show Tags

05 Jun 2005, 14:37
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

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?
Senior Manager
Joined: 17 May 2005
Posts: 271
Location: Auckland, New Zealand

### Show Tags

05 Jun 2005, 14:41
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

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?
Director
Joined: 18 Apr 2005
Posts: 545
Location: Canuckland

### Show Tags

05 Jun 2005, 14:48
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
Senior Manager
Joined: 17 May 2005
Posts: 271
Location: Auckland, New Zealand

### Show Tags

05 Jun 2005, 14:50
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!
Director
Joined: 01 Feb 2003
Posts: 844

### Show Tags

05 Jun 2005, 14:54
got it!
SVP
Joined: 05 Apr 2005
Posts: 1710

### Show Tags

05 Jun 2005, 19:11
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, 19:11
Display posts from previous: Sort by