ps: football - mgmat challenge : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 12:56

### 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

# ps: football - mgmat challenge

Author Message
Manager
Joined: 23 Dec 2006
Posts: 136
Followers: 1

Kudos [?]: 26 [0], given: 0

ps: football - mgmat challenge [#permalink]

### Show Tags

24 Sep 2007, 17:34
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Manhattan GMAT’s football team has 99 players. Each player has a uniform number from 1 to 99 and no two players share the same number. When football practice ends, all the players run off the field one-by-one in a completely random manner. What is the probability that the first four players off the field will leave in order of increasing uniform numbers (e.g., #2, then #6, then #67, then #72, etc)?

(A) 1/64
(B) 1/48
(C) 1/36
(D) 1/24
(E) 1/16
 Manhattan GMAT Discount Codes Magoosh Discount Codes Kaplan GMAT Prep Discount Codes
Intern
Joined: 14 Mar 2005
Posts: 28
Followers: 0

Kudos [?]: 8 [0], given: 0

### Show Tags

24 Sep 2007, 21:41
I think D...took me a while and i'm not sure if this is correct.

My reasoning is:

the size of the team is irrelevant. There are 4!=24 ways that the 4 jersey numbers of the first 4 players can be ordered. However since each number is distinct, there is only permutation in which the 4 numbers are in increasing order. Thus the probability is 1/24.
24 Sep 2007, 21:41
Display posts from previous: Sort by