|
Author |
Message |
|
TAGS:
|
|
|
Director
Joined: 01 Feb 2003
Posts: 897
Location: Hyderabad
Followers: 1
Kudos [?]:
9
[0], given: 0
|
Meg and Bob are among the 5 participants in a cycling race. [#permalink]
 05 Jun 2005, 14:11
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
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: 273
Location: Auckland, New Zealand
Followers: 1
Kudos [?]:
2
[0], given: 0
|
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
|
|
|
|
|
|
Director
Joined: 01 Feb 2003
Posts: 897
Location: Hyderabad
Followers: 1
Kudos [?]:
9
[0], given: 0
|
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?
|
|
|
|
|
|
Senior Manager
Joined: 17 May 2005
Posts: 273
Location: Auckland, New Zealand
Followers: 1
Kudos [?]:
2
[0], given: 0
|
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?
|
|
|
|
|
|
Director
Joined: 18 Apr 2005
Posts: 553
Location: Canuckland
Followers: 1
Kudos [?]:
3
[0], given: 0
|
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: 273
Location: Auckland, New Zealand
Followers: 1
Kudos [?]:
2
[0], given: 0
|
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: 897
Location: Hyderabad
Followers: 1
Kudos [?]:
9
[0], given: 0
|
 got it!
|
|
|
|
|
|
SVP
Joined: 05 Apr 2005
Posts: 1745
Followers: 2
Kudos [?]:
17
[0], given: 0
|
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.
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
close
