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.
May 0608:30 AM PDT
-09:30 AM PDT
In Episode 3 of our GMAT Ninja Critical Reasoning series, we tackle Discrepancy, Paradox, and Explain an Oddity questions. You know the feeling: the passage gives you two facts that seem completely contradictory....- May 08
08:00 AM PDT
-08:30 AM PDT
Join the special YouTube live-stream for selecting the winners of GMAT Club MBA Scholarships sponsored by Juno live. Watch who gets these coveted MBA scholarships offered by GMAT Club and Juno.
04 Mar 2019, 02:02
Kudos
Bookmarks
B
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:
87% (01:23) correct
13%
(02:33)
wrong
based on 103
sessions
History
Date
Time
Result
Not Attempted Yet
[GMAT math practice question]
How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?
A. 900
B. 1200
C. 1500
D. 1800
E. 2400
How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?
A. 900
B. 1200
C. 1500
D. 1800
E. 2400
04 Mar 2019, 04:05
Kudos
Bookmarks
We could almost rephrase this as three different questions:
How many ways to choose 2 from 4?
How many ways to choose 3 from 5?
How many ways to choose 3 from 6?
For this question, it's easier if you know the Combinations formula, but you could technically count.
Combinations formula = n! / k! (n-k)!
n = the bigger number (what we're choosing from)
k = the smaller number (how many we're choosing)
For example: how many ways to choose 2 from 4?
4! / 2! (4-2)!
4! / 2!2!
(4 x 3 x 2 x 1) / (2 x 1)(2 x 1)
24/4 = 6
But if we counted this and said the four males were ABCD, we could just list AB, AC, AD, BC, BD, and CD, and you can see it also equals 6.
Anyway, finding the other two:
How many ways to choose 3 from 5?
5! / 3! (5-3)!
5! / 3! 2!
We can cancel out 3! from both numerator and denominator.
5 x 4 / 2
20/2 = 10
How many ways to choose 3 from 6?
6! / 3! (6-3)!
6! / 3! 3!
We can cancel out one 3! from both numerator and denominator.
6 x 5 x 4 / 3 x 2
120 / 6 = 20
Back to our original questions:
How many ways to choose 2 from 4? 6
How many ways to choose 3 from 5? 10
How many ways to choose 3 from 6? 20
Now, multiply all those numbers together!
6 x 10 x 20 = 1200
How many ways to choose 2 from 4?
How many ways to choose 3 from 5?
How many ways to choose 3 from 6?
For this question, it's easier if you know the Combinations formula, but you could technically count.
Combinations formula = n! / k! (n-k)!
n = the bigger number (what we're choosing from)
k = the smaller number (how many we're choosing)
For example: how many ways to choose 2 from 4?
4! / 2! (4-2)!
4! / 2!2!
(4 x 3 x 2 x 1) / (2 x 1)(2 x 1)
24/4 = 6
But if we counted this and said the four males were ABCD, we could just list AB, AC, AD, BC, BD, and CD, and you can see it also equals 6.
Anyway, finding the other two:
How many ways to choose 3 from 5?
5! / 3! (5-3)!
5! / 3! 2!
We can cancel out 3! from both numerator and denominator.
5 x 4 / 2
20/2 = 10
How many ways to choose 3 from 6?
6! / 3! (6-3)!
6! / 3! 3!
We can cancel out one 3! from both numerator and denominator.
6 x 5 x 4 / 3 x 2
120 / 6 = 20
Back to our original questions:
How many ways to choose 2 from 4? 6
How many ways to choose 3 from 5? 10
How many ways to choose 3 from 6? 20
Now, multiply all those numbers together!
6 x 10 x 20 = 1200
