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.
15 Feb 2021, 09:00
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
72% (02:34) correct
28%
(02:34)
wrong
based on 138
sessions
History
Date
Time
Result
Not Attempted Yet
A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. If internal arrangement inside the car does not matter then the number of ways in which they can travel, is
(A) 91
(B) 122
(C) 126
(D) 3920
(E) 4120
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 91
(B) 122
(C) 126
(D) 3920
(E) 4120
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
15 Feb 2021, 10:22
Kudos
Bookmarks
Bunuel
So we have two cars, and ways in which the 8 students can be made to sit depending on the capacity of cars.
(1) 5 and 3
We can choose 5 out of 8, and remaining 3 will go to the other car. => 8C5=\(\frac{8*7*6}{3*2}=8*7=56\)
(2) 4 and 4
We can choose 4 out of 8, and remaining 4 will go to the other car. => 8C4=\(\frac{8*7*6*5}{4*3*2}=7*2*5=70\)
Total =56+70=126
C
09 Mar 2021, 14:16
Kudos
Bookmarks
Bunuel
Solution:
Let’s first assume that the students are indistinguishable. Then the problem becomes: How many ways can 2 positive integers (where the first addend can’t be more than 4 and the second can’t be more than 5) add up to 8? We can see that we can have:
3 + 5 = 8 and 4 + 4 = 8
Of course, the students (and in general, people) are distinguishable, so we have to consider each of the two cases above:
Case 1: 3 + 5 = 8
That is, the first car gets 3 students and the second 5 students. Since the students are distinguishable, the number of choices the first car has is 8C3 and the number of choices the second car has is 5C5 (after 3 students have been chosen to sit in the first car). Therefore, the number of ways the 8 students can be distributed into 2 cars in this case is 8C3 x 5C5 = (8 x 7 x 6)/(3 x 2) x 1 = 8 x 7 = 56.
Case 2: 4 + 4 = 8
That is, the first car gets 4 students and the second 4 students. Since the students are distinguishable, the number of choices the first car has is 8C4 and the number of choices the second car has is 4C4 (after 4 students have been chosen to sit in the first car). Therefore, the number of ways the 8 students can be distributed into 2 cars in this case is 8C4 x 4C4 = (8 x 7 x 6 x 5)/(4 x 3 x 2) x 1 = 2 x 7 x 5 = 70.
Thus, the total number of ways in both cases is 56 + 70 = 126.
Answer: C
