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

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VP
Joined: 22 Nov 2007
Posts: 1063
Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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07 Jan 2008, 06:05
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64% (00:59) correct 36% (01:35) wrong based on 1248 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 47037
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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26 Feb 2012, 23:01
5
13
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.

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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Manager
Joined: 01 Sep 2007
Posts: 95
Location: Astana
Re: still on perms and combs [#permalink]

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08 Jan 2008, 19:02
13
3
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
##### General Discussion
Director
Joined: 12 Jul 2007
Posts: 854
Re: still on perms and combs [#permalink]

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07 Jan 2008, 06:13
5
8
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
VP
Joined: 28 Dec 2005
Posts: 1482
Re: still on perms and combs [#permalink]

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08 Jan 2008, 18:18
1
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 ?
Director
Joined: 09 Aug 2006
Posts: 746
Re: still on perms and combs [#permalink]

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09 Jan 2008, 02:16
4
6
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

SVP
Joined: 29 Mar 2007
Posts: 2491
Re: still on perms and combs [#permalink]

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09 Jan 2008, 21:11
1
3
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.
Manager
Joined: 27 Oct 2008
Posts: 181
Re: still on perms and combs [#permalink]

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27 Sep 2009, 10:44
1
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
Intern
Joined: 24 Feb 2012
Posts: 31
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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26 Feb 2012, 19:27
6
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...
Math Expert
Joined: 02 Sep 2009
Posts: 47037
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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26 Feb 2012, 23:04
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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Joined: 16 Oct 2010
Posts: 8124
Location: Pune, India
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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27 Feb 2012, 05:07
1
1
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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Contact: bansal.karishma@gmail.com

Manager
Joined: 02 Apr 2012
Posts: 76
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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06 Jul 2013, 19:03
1
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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Joined: 06 Sep 2013
Posts: 1869
Concentration: Finance
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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30 Dec 2013, 06:09
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
Intern
Joined: 26 Mar 2013
Posts: 21
Location: India
Concentration: Finance, Strategy
Schools: Booth PT '18 (S)
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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19 Jan 2015, 01:08
If u catch that its simple symmetry ans takes < 30 sec

total prob = 5! = 120

the chances of one being ahead of another are equal....so if we are looking for just one to be ahead of the other => half => 120/2 =60
Intern
Joined: 20 Dec 2014
Posts: 21
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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12 Mar 2015, 11:34
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

When I first saw this I thought combination but then realized its permutation because order does matter(Meg has to finish ahead of Bob). So we can use n!/(n-k)! n represents the 5 participants and k would be the 3 remaining cyclers which the orders do not matter correct? so 5!/(5-3)!= 60 answer is C

Is this a correct approach?
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Joined: 16 Oct 2010
Posts: 8124
Location: Pune, India
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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12 Mar 2015, 21:46
GMAT01 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

When I first saw this I thought combination but then realized its permutation because order does matter(Meg has to finish ahead of Bob). So we can use n!/(n-k)! n represents the 5 participants and k would be the 3 remaining cyclers which the orders do not matter correct? so 5!/(5-3)!= 60 answer is C

Is this a correct approach?

I don't think so. You have used nPr = 5P3 which we use when we have to choose and arrange 3 objects out of 5. The logic here is a bit different. I suggest you to check out the solutions given above.
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Intern
Joined: 19 Mar 2015
Posts: 1
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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13 May 2015, 01:50
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

My approach was I counted Meg and Bob as one entity (in which Meg is always ahead of Bob) and then carried out the Permutations for 4 participants instead of 5. That way 4 participants will be arranged in 4! factorial ways ie. 24 ways. But when I checked the answers I found out that there was some flaw in my logic. Can someone please let me know what went wrong ?
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Joined: 04 Jan 2015
Posts: 1755
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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13 May 2015, 02:25
2
HarshitBShah wrote:
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

My approach was I counted Meg and Bob as one entity (in which Meg is always ahead of Bob) and then carried out the Permutations for 4 participants instead of 5. That way 4 participants will be arranged in 4! factorial ways ie. 24 ways. But when I checked the answers I found out that there was some flaw in my logic. Can someone please let me know what went wrong ?

Hi HarshitBShah,

You have considered one of the scenarios of the total possible combinations. When you assume Meg & Bob as one entity you only take the case where Bob is the next person behind Meg.

Imagine a situation where Meg finishes 1st, since Bob has to finish behind Meg, Bob can finish at any place from 2nd to the 5th. Similarly when Meg finishes 2nd, Bob can finish at any place from 3rd to 5th and so on. Combining them as one entity would have worked had the question constrained Bob & Meg to finish the race in consecutive positions.

Hope its clear

Regards
Harsh
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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13 May 2015, 22:06
HarshitBShah wrote:
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

My approach was I counted Meg and Bob as one entity (in which Meg is always ahead of Bob) and then carried out the Permutations for 4 participants instead of 5. That way 4 participants will be arranged in 4! factorial ways ie. 24 ways. But when I checked the answers I found out that there was some flaw in my logic. Can someone please let me know what went wrong ?

"Meg finishes ahead of Bob" is different from "Bob finishes immediately after Meg".

You accounted for cases such as:
Meg, Bob, A, B, C
and
C, Meg, Bob, A, B

etc but how about cases such as:
Meg, B, Bob, A, C?

You need to use the principle of symmetry here. Check out the link in gave in my post above.
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Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

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24 Aug 2016, 12:12
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

this is similar to the problem, "How many 5 letter words can we form with a 5 letter word in which 2 letters are repeated, ex. ALPHA"
and the answer is $$\frac{5!}{2!}$$ = 60.
Meg following Bob can be treated as one group or identical and it will be exactly $$\frac{1}{2}$$ times of Total and equal to Bob following Meg.
Re: Meg and Bob are among the 5 participants in a cycling race.   [#permalink] 24 Aug 2016, 12:12

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