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

It is currently 22 Sep 2014, 22:22

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

counting methods

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Intern
Intern
avatar
Joined: 07 Aug 2007
Posts: 16
Followers: 0

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

counting methods [#permalink] New post 11 Nov 2007, 20:23
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
there are 5 cars to be displayed in 5 paring spaces with all the cars facing the same direction. of the 5 cars, 3 are red, 1 blue and 1 yellow. if the cars are identical except for the color, how many different display arrangements of the 5 cars are possible?

20
25
40
60
125...
Manager
Manager
avatar
Joined: 25 Jul 2007
Posts: 108
Followers: 1

Kudos [?]: 12 [0], given: 0

 [#permalink] New post 11 Nov 2007, 20:37
Option A is the answer

Ignore the red cars. We will concentrate on the yellow and the blue car.
There are 5 possible positions for the yellow car. For each such position of the yellow car, there are 4 possible positions for the blue car. Therefore the answer is 5*4 =20.
Director
Director
avatar
Joined: 09 Aug 2006
Posts: 767
Followers: 1

Kudos [?]: 49 [0], given: 0

GMAT Tests User
 [#permalink] New post 12 Nov 2007, 00:32
jbs wrote:
Option A is the answer

Ignore the red cars. We will concentrate on the yellow and the blue car.
There are 5 possible positions for the yellow car. For each such position of the yellow car, there are 4 possible positions for the blue car. Therefore the answer is 5*4 =20.


Can you pls. explain why you ignore the number of red cars?
Manager
Manager
avatar
Joined: 25 Jul 2007
Posts: 108
Followers: 1

Kudos [?]: 12 [0], given: 0

 [#permalink] New post 12 Nov 2007, 03:14
GK_Gmat wrote:
jbs wrote:
Option A is the answer

Ignore the red cars. We will concentrate on the yellow and the blue car.
There are 5 possible positions for the yellow car. For each such position of the yellow car, there are 4 possible positions for the blue car. Therefore the answer is 5*4 =20.


Can you pls. explain why you ignore the number of red cars?


They say that the cars are identical in every way except for the colour.

Therefore, once you have decided on the positions of the yellow and the blue car, the 3 red cars can be arranged in any manner among themselves to give the same display arrangement.

Here's an example. For the sake of understanding, assume that the 3 red cars are r1,r2 and r3.

here's a possible display arrangement

Y B r1 r2 r3

now note that the above display arrangement will look exactly the same as

Y B r2 r3 r1

My understanding of the question tells me that the exact position of a particular car is not important but the display arrangement is.

Therefore, as we seek to find the number of different display arrangements, we can ignore the similar red cars.

Also, refer to GMAT TIGER's post for a more conceptual approach.
CEO
CEO
User avatar
Joined: 29 Aug 2007
Posts: 2501
Followers: 53

Kudos [?]: 507 [0], given: 19

GMAT Tests User
 [#permalink] New post 12 Nov 2007, 11:36
jbs wrote:
GK_Gmat wrote:
jbs wrote:
Option A is the answer

Ignore the red cars. We will concentrate on the yellow and the blue car.
There are 5 possible positions for the yellow car. For each such position of the yellow car, there are 4 possible positions for the blue car. Therefore the answer is 5*4 =20.


Can you pls. explain why you ignore the number of red cars?


They say that the cars are identical in every way except for the colour.

Therefore, once you have decided on the positions of the yellow and the blue car, the 3 red cars can be arranged in any manner among themselves to give the same display arrangement.

Here's an example. For the sake of understanding, assume that the 3 red cars are r1,r2 and r3.

here's a possible display arrangement

Y B r1 r2 r3

now note that the above display arrangement will look exactly the same as

Y B r2 r3 r1

My understanding of the question tells me that the exact position of a particular car is not important but the display arrangement is.

Therefore, as we seek to find the number of different display arrangements, we can ignore the similar red cars.

Also, refer to GMAT TIGER's post for a more conceptual approach.


perfect. in repetitive case,

req possibilities = total possibilities/repetitive possibilities
req possibilities = 5!/3!


because r1 r2 r3 and r3 r2 r1 are same and counted as 1 possibility. so we need to divide 5! by 3!.
  [#permalink] 12 Nov 2007, 11:36
    Similar topics Author Replies Last post
Similar
Topics:
Counting rxs0005 2 20 Aug 2010, 14:58
counting sanjay_gmat 4 03 Apr 2009, 13:25
Counting Shelleb17 6 17 Jan 2009, 04:32
PS (Counting Methods) netcaesar 2 13 Dec 2008, 10:04
Counting Methods - GMATPrep jp888 1 31 Aug 2007, 09:23
Display posts from previous: Sort by

counting methods

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

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®.