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

It is currently 31 Aug 2015, 09:47
GMAT Club Tests

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

As part of a game, four people each must secretly choose an

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
VP
VP
User avatar
Joined: 25 Nov 2004
Posts: 1494
Followers: 6

Kudos [?]: 50 [0], given: 0

As part of a game, four people each must secretly choose an [#permalink] New post 21 Apr 2005, 22:45
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?
Manager
Manager
avatar
Joined: 05 Feb 2005
Posts: 116
Location: San Jose
Followers: 1

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

 [#permalink] New post 21 Apr 2005, 23:12
As part of a game, four people each must secretly choose an integer between 1 and 4, inclusive. What is the approximate likelihood that all four people will choose different numbers?

Total number of choice combos = 4x4x4x4
Number of choice combos where they choose diff numbers = 4!
(arrange 4 numbers into 4 empty slots = 4P4 = 4!)

Hence the answer = 4! / 256 = 3/32
_________________

Anyone who has never made a mistake has never tried anything new. -Albert Einstein.

VP
VP
User avatar
Joined: 25 Nov 2004
Posts: 1494
Followers: 6

Kudos [?]: 50 [0], given: 0

 [#permalink] New post 25 Apr 2005, 07:46
no more..................... :oops:
VP
VP
avatar
Joined: 30 Sep 2004
Posts: 1488
Location: Germany
Followers: 4

Kudos [?]: 131 [0], given: 0

 [#permalink] New post 25 Apr 2005, 11:08
MA wrote:
no more..................... :oops:


number of different arrangements is 4! and total of 4^4 is possible. so it is 4!/4^4 = 3/32
:-D
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

VP
VP
User avatar
Joined: 13 Jun 2004
Posts: 1119
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 33 [0], given: 0

 [#permalink] New post 25 Apr 2005, 18:36
Same approach as mckenna

total possibilities : 4^4
possibilities of 4 different numbers : 4*3*2*1 = 24

prob : 24/4^4 = 6/4^3 = 3/32
  [#permalink] 25 Apr 2005, 18:36
Display posts from previous: Sort by

As part of a game, four people each must secretly choose an

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

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®.