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

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Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 07 Jan 2008, 05:05
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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
[Reveal] Spoiler: OA
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Re: still on perms and combs [#permalink] New post 07 Jan 2008, 05:13
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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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Re: still on perms and combs [#permalink] New post 08 Jan 2008, 17:18
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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Re: still on perms and combs [#permalink] New post 08 Jan 2008, 18:02
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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] New post 09 Jan 2008, 01:16
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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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Re: still on perms and combs [#permalink] New post 09 Jan 2008, 20:11
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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Re: still on perms and combs [#permalink] New post 27 Sep 2009, 09:44
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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

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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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 26 Feb 2012, 18:27
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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] New post 26 Feb 2012, 22:01
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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

Total # of ways the race can be finished is 5!. In half of the cases Meg finishes ahead of Bob and in other half Bob finishes ahead of Meg. So, ways Meg to finish ahead of Bob is 5!/2=60.

Answer: C.


fortsill wrote:
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...


I was able to recall several such questions from combinations:
mother-mary-comes-to-me-86407.html (or: mary-and-joe-126407.html);
six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
in-how-many-different-ways-can-the-letters-a-a-b-91460.html
goldenrod-and-no-hope-are-in-a-horse-race-with-6-contestants-82214.html

Check our question banks for many more questions on combinations.
PS: search.php?search_id=tag&tag_id=52
DS: search.php?search_id=tag&tag_id=31

Hope it helps
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 26 Feb 2012, 22:04
Expert's post
Also try Combinatorics chapter of Math Book to have an idea about the staff that is tested on the GMAT: math-combinatorics-87345.html

Finally try Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html

Hope it helps.
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 27 Feb 2012, 04:07
Expert's post
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. [#permalink] New post 06 Jul 2013, 18:03
Subject:
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?

I solved in a different way:
All the possible positions: 5! = 120
Restriction: we have to eliminate all the combinations in which Bob finishes ahead of Meg:

If Bob finishes 1st, (_ _ _ _ B) Meg could finish 2nd, 3th, 4th or 5th --> 4 possibilities.
If Bob finishes 2nd, Meg finishes 3th, 4th or 5th. --> 3 possibilities
The same for Bob 3th and 4th.
Bob can not finishes 5th.

This sum up to 10 combinatios: 4 + 3 + 2 + 1 = 10.

In addition, we have the rest of the combinations of the of the positions remaining 3 competitors: 3!

Now, we have to combine the positions of Meg and Bob with the positions of the rest of the competitors:

3! * 10 = 60

Finally:
120 - 60 = 60
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] New post 30 Dec 2013, 05:09
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


By symmetry 5! /2

Meg will finish ahead of Bob in half of all the possible ways

Hope it helps
Cheers!

J :)
Re: Meg and Bob are among the 5 participants in a cycling race.   [#permalink] 30 Dec 2013, 05:09
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