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VP
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Meg and Bob are among the 5 participants in a cycling race. [#permalink]
 07 Jan 2008, 06:05
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Question Stats:
65% (01:45) correct
34% (01:16) wrong based on 1 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
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Director
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Re: still on perms and combs [#permalink]
07 Jan 2008, 06:13
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There are 5! possible ways for the 5 participants to finish the race without any ties.
Logically, Meg will finish ahead of Bob in exactly half of these outcomes
5!/2 = 60
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SVP
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Re: still on perms and combs [#permalink]
08 Jan 2008, 18:18
ok, this is how i started to think about it ... didnt get the right answer though
there are 5 spots, so meg could come in first, second, third, fourth or fifth
if she comes in first, then there are 4 spots behind her that bob could be in ... so thats 4! = 24 ways
if she comes in second, then there are 3 spots behind her ... bob could be in any of those, so thats 3! = 6 ways
if she was in third, only two spots behind her, so thats 2 ways
if she comes in fourth, only one way ... adding them up, i get 24+6+2+1 = 33 ways.
Where did i go wrong ?
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Manager
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Re: still on perms and combs [#permalink]
08 Jan 2008, 19:02
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pmenon wrote: ok, this is how i started to think about it ... didnt get the right answer though
there are 5 spots, so meg could come in first, second, third, fourth or fifth
if she comes in first, then there are 4 spots behind her that bob could be in ... so thats 4! = 24 ways
if she comes in second, then there are 3 spots behind her ... bob could be in any of those, so thats 3! = 6 ways
if she was in third, only two spots behind her, so thats 2 ways
if she comes in fourth, only one way ... adding them up, i get 24+6+2+1 = 33 ways.
Where did i go wrong ? look at the problem this way: M _ _ _ _ = 4! ways as you said _ M _ _ _ = 3 ways for Bob, still 3 options for the next (because unlike Bob she can occupy the first slot), 2 options for the the fourth one and 1 for the last. So 3*3*2*1 which is 3*3! _ _ M _ _ = 2*3! _ _ _ M _ = 1*3! 24 + 18 + 12 + 6 = 60
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Re: still on perms and combs [#permalink]
09 Jan 2008, 02:16
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marcodonzelli 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 C. M finishes first then B can finish 4 ways and other three can finish in 3! ways = 1*4*3! = 24 M finishes second then B can finish in 3 ways and other three can finish in 3! ways = 1*3*3! = 18 M finishes third then B can finish in 2 ways and other three can finish in 3! ways = 1*2*3! = 12 M finishes fourth then B can finish in 1 way and other three can finish in 3! ways = 1*1*3! = 6 Answer = 24+18+12+6 = 60
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CEO
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Re: still on perms and combs [#permalink]
09 Jan 2008, 21:11
marcodonzelli 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 MBXXX we have 4! possibilities = 24 XMBXX we have 3! ways to arrange the X's and 3 ways to arrange B so 3!*3 =18 XXMBX we have 3! ways to arrange the X's and 2 ways to arrange B. so 3!*2 =12 XXXMB only 3! =6 24+18+12+6 = 60.
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Manager
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Re: still on perms and combs [#permalink]
27 Sep 2009, 10:44
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
Ans: C. 60
M stands for Meg
If M takes first place, the other 4 can be arranged in 4! ways.
If M takes second place, the other 4 can be arranged in 3 * 3! ways such that Bob finishes after Meg.
If M takes third place, the other 4 can be arranged in 6 * 2! ways such that Bob finishes after Meg.
If M takes fourth place, the other 4 can be arranged in 3! ways such that Bob finishes after Meg.
M cannot take fifth place, because in tat case Bob will finish ahead of Meg.
So total number of possibilities is
= 4! + 3 * 3! + 6 * 2! + 3!
= 60
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Intern
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]
26 Feb 2012, 19:27
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hmm, I went about it the long way until I read the response here.
When M finishes 1st:
Permutations ---> (4P3) = 4! = 24
When M finishes 2nd:
Permutations ---> (3P1 * 3P3) = 3 * 3! = 18
(3P1 ways of arranging people before M, 3P3 ways of arranging people after M)
When M finishes 3rd:
Permutations ---> (3P2 * 2P2) = 6 * 2! = 12
(3P2 ways of arranging people before M, 2P2 ways of arranging people after M)
When M finishes 4th:
Permutations ---> (3P3 * 1P1) = 3! * 1 = 6
(3P3 ways of arranging people before M, 1P1 ways of arranging people after M)
Total = 24 + 18 + 12 + 6 = 60.
That being said, the other technique of arriving at the same solution by 5!/2 is far more elegant and time saving!
Do you know of any other such examples where the problem seems a lot simpler than it really is...
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]
26 Feb 2012, 23:01
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]
26 Feb 2012, 23:04
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Veritas Prep GMAT Instructor
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]
27 Feb 2012, 05:07
fortsill wrote: hmm, I went about it the long way until I read the response here.
What happened with you, happens with many other people. I have given a detailed explanation of questions of this type in this post: http://www.veritasprep.com/blog/2011/10 ... s-part-ii/
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Re: Meg and Bob are among the 5 participants in a cycling race.
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27 Feb 2012, 05:07
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