Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 May 2016, 05:23

### 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.

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 9

Kudos [?]: 374 [2] , given: 0

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

### Show Tags

07 Jan 2008, 06:05
2
KUDOS
24
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

60% (01:59) correct 40% (01:17) wrong based on 785 sessions

### HideShow timer Statictics

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
Director
Joined: 12 Jul 2007
Posts: 862
Followers: 13

Kudos [?]: 244 [3] , given: 0

Re: still on perms and combs [#permalink]

### Show Tags

07 Jan 2008, 06:13
3
KUDOS
4
This post was
BOOKMARKED
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
SVP
Joined: 28 Dec 2005
Posts: 1575
Followers: 3

Kudos [?]: 117 [0], given: 2

Re: still on perms and combs [#permalink]

### Show Tags

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 ?
Manager
Joined: 01 Sep 2007
Posts: 100
Location: Astana
Followers: 1

Kudos [?]: 21 [7] , given: 0

Re: still on perms and combs [#permalink]

### Show Tags

08 Jan 2008, 19:02
7
KUDOS
1
This post was
BOOKMARKED
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
Director
Joined: 09 Aug 2006
Posts: 763
Followers: 1

Kudos [?]: 132 [3] , given: 0

Re: still on perms and combs [#permalink]

### Show Tags

09 Jan 2008, 02:16
3
KUDOS
4
This post was
BOOKMARKED
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
CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 18

Kudos [?]: 320 [0], given: 0

Re: still on perms and combs [#permalink]

### Show Tags

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.
Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 1

Kudos [?]: 126 [1] , given: 3

Re: still on perms and combs [#permalink]

### Show Tags

27 Sep 2009, 10:44
1
KUDOS
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: 33
Followers: 0

Kudos [?]: 13 [4] , given: 18

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

### Show Tags

26 Feb 2012, 19:27
4
KUDOS
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: 32610
Followers: 5651

Kudos [?]: 68585 [5] , given: 9815

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

### Show Tags

26 Feb 2012, 23:01
5
KUDOS
Expert's post
8
This post was
BOOKMARKED
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
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 32610
Followers: 5651

Kudos [?]: 68585 [0], given: 9815

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

### Show Tags

26 Feb 2012, 23:04
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.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6482
Location: Pune, India
Followers: 1760

Kudos [?]: 10503 [1] , given: 206

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

### Show Tags

27 Feb 2012, 05:07
1
KUDOS
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/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Joined: 02 Apr 2012 Posts: 76 Location: Argentina Concentration: Entrepreneurship, Finance GMAT 1: 680 Q49 V34 WE: Consulting (Consulting) Followers: 1 Kudos [?]: 40 [0], given: 155 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 06 Jul 2013, 19:03 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 _________________ Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 43 Kudos [?]: 456 [0], given: 355 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 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 Senior Manager Joined: 10 Mar 2013 Posts: 290 GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Followers: 3 Kudos [?]: 71 [0], given: 2404 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 05 Oct 2014, 17:17 VeritasPrepKarishma wrote: 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/ Helped me! Thanks! Intern Joined: 26 Mar 2013 Posts: 23 Location: India Concentration: Finance, Strategy Schools: Booth PT '18 (S) Followers: 0 Kudos [?]: 8 [0], given: 2 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 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: 15 Followers: 0 Kudos [?]: 1 [0], given: 23 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 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? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6482 Location: Pune, India Followers: 1760 Kudos [?]: 10503 [0], given: 206 Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink] ### Show Tags 12 Mar 2015, 21:46 Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Intern
Joined: 19 Mar 2015
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 89

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

### Show Tags

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 ?
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 338
Followers: 87

Kudos [?]: 699 [1] , given: 84

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

### Show Tags

13 May 2015, 02:25
1
KUDOS
Expert's post
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
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6482
Location: Pune, India
Followers: 1760

Kudos [?]: 10503 [0], given: 206

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

### Show Tags

13 May 2015, 22:06
Expert's post
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".
Meg could finish way ahead of Bob or just ahead.

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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Re: Meg and Bob are among the 5 participants in a cycling race.   [#permalink] 13 May 2015, 22:06
Similar topics Replies Last post
Similar
Topics:
12 Anne and Beth will participate in a sack race (In a sack race, people 4 04 Oct 2014, 00:59
Tough and tricky 5: Race 5 11 Oct 2009, 18:14
1 5 women are in a race. What is the probability that 10 19 Nov 2007, 00:09
12 Meg and Bob are among the 5 participants in a cycling race. 14 22 Oct 2007, 20:14
8 Meg and Bob are among the 5 participants in a cycling race. 8 23 Oct 2006, 10:43
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.