It is currently 18 Nov 2017, 18:39

### 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.

Author Message
TAGS:

### Hide Tags

Manager
Joined: 07 Jun 2006
Posts: 109

Kudos [?]: 34 [0], given: 0

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

### Show Tags

23 Oct 2006, 10:43
3
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

64% (01:42) correct 36% (01:04) wrong based on 215 sessions

### HideShow timer Statistics

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Aug 2014, 01:40, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

Kudos [?]: 34 [0], given: 0

Manager
Joined: 07 Jun 2006
Posts: 109

Kudos [?]: 34 [0], given: 0

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

### Show Tags

23 Oct 2006, 10:46
I got 33 here

24+6+2+1, is it rite?

Kudos [?]: 34 [0], given: 0

VP
Joined: 15 Jul 2004
Posts: 1438

Kudos [?]: 225 [0], given: 13

Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

### Show Tags

23 Oct 2006, 10:56
girikorat wrote:
If Bob and Jen are two of 5 participants in a race, how many different ways can the race finish where Jen always finishes in front of Bob?

It should be 4! = 24. Basically since Jen and Bob's positions are fixed - we can just treat them as ONE combination. Taking the remaining three folks + ONE combination of Jen and Bob you have 4 permutations for the race to finish.

Kudos [?]: 225 [0], given: 13

Manager
Joined: 04 Jan 2006
Posts: 58

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

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

### Show Tags

23 Oct 2006, 11:15
1
This post was
BOOKMARKED
Is it 60?
When Jen is in posion 1:
4! ways to arrange the other 4.

When Jim is in pos 2:
Bob has to be in 3rd, 4th or 5th pos. And the other 3 can be positioned in 3! ways
So 3x3!

When Jen is in pos 3:
Bob has to be in 4th ot 5th. Others can be positioned in 3! ways
so 2x3!

When Jen is in 4th:
Bob has to the 5th person. Others- 3!

So total = 4! + 3x3! + 2x3! + 3!
= 4! + 3!x6 = 24 + 36 = 60

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

Manager
Joined: 07 Jun 2006
Posts: 109

Kudos [?]: 34 [0], given: 0

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

### Show Tags

23 Oct 2006, 11:20
yeah its 60....I think I got it now....

its 4*3!+3*3!+2*3!+1*3! = 60

Thanks guys

Kudos [?]: 34 [0], given: 0

Tutor
Joined: 20 Apr 2012
Posts: 100

Kudos [?]: 338 [3], given: 36

Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

### Show Tags

08 Aug 2014, 01:03
3
KUDOS
1
This post was
BOOKMARKED
Race can finish in 5!=120 ways, and exactly half of them Jen always finishes in front of Bob. The correct answer is 60.
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

Last edited by smyarga on 08 Aug 2014, 13:31, edited 1 time in total.

Kudos [?]: 338 [3], given: 36

Intern
Joined: 04 Feb 2014
Posts: 6

Kudos [?]: [0], given: 14

Schools: Tepper MISM '18
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

### Show Tags

08 Aug 2014, 12:54
smyarga wrote:
Race can finish in 6!=120 ways, and exactly half of them Jen always finishes in front of Bob. The correct answer is 60.

Think you mean 5! = 120

Kudos [?]: [0], given: 14

Tutor
Joined: 20 Apr 2012
Posts: 100

Kudos [?]: 338 [1], given: 36

Location: Ukraine
GMAT 1: 690 Q51 V31
GMAT 2: 730 Q51 V38
WE: Education (Education)
Re: Meg and Bob are among the 5 participants in a cycling race. [#permalink]

### Show Tags

08 Aug 2014, 13:32
1
KUDOS
sismail wrote:
smyarga wrote:
Race can finish in 6!=120 ways, and exactly half of them Jen always finishes in front of Bob. The correct answer is 60.

Think you mean 5! = 120

Yep:) edited) thanks!
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

Kudos [?]: 338 [1], given: 36

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132619 [1], given: 12326

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

### Show Tags

12 Aug 2014, 01:41
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
girikorat 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

OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html

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.

Similar questions to practice:
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
_________________

Kudos [?]: 132619 [1], given: 12326

Re: Meg and Bob are among the 5 participants in a cycling race.   [#permalink] 12 Aug 2014, 01:41
Display posts from previous: Sort by