Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
84% (00:38) correct
16%
(00:49)
wrong
based on 1096
sessions
History
Date
Time
Result
Not Attempted Yet
There are 10 children in a company's day-care center, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?
A) 100
B) 90
C) 50
D) 45
E) 25
A) 100
B) 90
C) 50
D) 45
E) 25
12 Nov 2008, 06:44
Kudos
Bookmarks
First way
1. first child out of 10 - 10 possibilities
2. second child out of 9 - 9 possibilities. the total number of pairs = 9*10=90
3. exclude xy yx cases: 90/2=45
Second way
N=10C2=10!/(8!2!)=10*9/2=45
1. first child out of 10 - 10 possibilities
2. second child out of 9 - 9 possibilities. the total number of pairs = 9*10=90
3. exclude xy yx cases: 90/2=45
Second way
N=10C2=10!/(8!2!)=10*9/2=45
03 Mar 2015, 22:49
Kudos
Bookmarks
Hi All,
This is an example of a fairly-straightforward Combinations question. It asks how many different pairs of children (meaning "sets of 2") are there that can be selected from a group of 10 children. In these situations, the ORDER of the two children does not matter, so if 'child A' plays 'child B' then that pair is the SAME as when 'child B' plays 'child A.' As such, you are NOT supposed to count that pairing twice. The Combinations Formula removes all of those duplicate pairs.
Combinations = N!/[K!(N-K)!] Where N is the total number of children and K is the number that you are going to pick.
Here, we have N = 10 and K = 2
10!/[2!8!] = (10)(9)/(2)(1) = 45 different pairs of 2 children.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
This is an example of a fairly-straightforward Combinations question. It asks how many different pairs of children (meaning "sets of 2") are there that can be selected from a group of 10 children. In these situations, the ORDER of the two children does not matter, so if 'child A' plays 'child B' then that pair is the SAME as when 'child B' plays 'child A.' As such, you are NOT supposed to count that pairing twice. The Combinations Formula removes all of those duplicate pairs.
Combinations = N!/[K!(N-K)!] Where N is the total number of children and K is the number that you are going to pick.
Here, we have N = 10 and K = 2
10!/[2!8!] = (10)(9)/(2)(1) = 45 different pairs of 2 children.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
