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

 It is currently 07 Mar 2014, 07:40

### 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

# If Bob and Jen are two of 5 participants in a race, how many

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Manager
Joined: 07 Jun 2006
Posts: 113
Followers: 2

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

If Bob and Jen are two of 5 participants in a race, how many [#permalink]  23 Oct 2006, 09:43
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?
Manager
Joined: 07 Jun 2006
Posts: 113
Followers: 2

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

[#permalink]  23 Oct 2006, 09:46
I got 33 here

24+6+2+1, is it rite?
VP
Joined: 15 Jul 2004
Posts: 1474
Schools: Wharton (R2 - submitted); HBS (R2 - submitted); IIMA (admitted for 1 year PGPX)
Followers: 14

Kudos [?]: 79 [0], given: 13

Re: Race [#permalink]  23 Oct 2006, 09:56
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.
Manager
Joined: 04 Jan 2006
Posts: 58
Followers: 1

Kudos [?]: 1 [0], given: 0

[#permalink]  23 Oct 2006, 10:15
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
Manager
Joined: 07 Jun 2006
Posts: 113
Followers: 2

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

[#permalink]  23 Oct 2006, 10:20
yeah its 60....I think I got it now....

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

Thanks guys
[#permalink] 23 Oct 2006, 10:20
Similar topics Replies Last post
Similar
Topics:
Meg and Bob are among the 5 participants in a cycling race. 7 05 Jun 2005, 13:11
Meg and Bob are among the 5 participants in a cycling race. 5 10 Dec 2005, 15:22
Q17: Meg and Bob are among the 5 participants in a cycling 2 03 Jun 2007, 19:40
3 If Bob and Jen are two of 5 participants in a race, how many 12 22 Oct 2007, 19:14
14 Meg and Bob are among the 5 participants in a cycling race. 12 07 Jan 2008, 05:05
Display posts from previous: Sort by

# If Bob and Jen are two of 5 participants in a race, how many

 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®.