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.
17 Feb 2021, 07:00
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
58% (02:34) correct
42%
(02:12)
wrong
based on 146
sessions
History
Date
Time
Result
Not Attempted Yet
Number of ways in which four different toys and five indistinguishable marbles can be distributed between Amar, Akbar and Anthony, if each child receives at least one toy and one marble, is
(A) 42
(B) 100
(C) 150
(D) 216
(E) 226
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 42
(B) 100
(C) 150
(D) 216
(E) 226
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
17 Feb 2021, 17:37
Kudos
Bookmarks
I’ll give it a try....
I believe the question is saying “at least one toy and at least one marble”
(1st) distributing the 4 different toys to 3 different people
Each person must receive at least 1 of the 4 toys. This means one person will receive 2 toys and the other two people will receive just 1 toy.
[2 ; 1 ; 1]
Step 1: distribute the different toys into identical stacks as listed above
“4 choose 2” * “2 choose 1” * 1 choose 1” = 4! / (2! * 1! * 1!)
However, since there are identical amounts of toys placed in the stacks that contain 1, we will have repeat counts. To remove the overcounting, we need to divide by 2!
End up with: 4! / (2! * 2!*) = 6 ways
AND
Step 2: Because we have 3 distinct groups of people, we need to arrange the “stacks” among the 3 people. Because each stack has different items, it matters which stack with 1 toy goes to which person.
We can arrange the 3 different stacks among 3 different people in 3! Ways
Toys can be distributed in: 6 * 3! = 36 ways
And
(2nd) distribute 5 identical marbles among 3 distinct “groups” (people)
Letting the 3 different people be variables A B and C
A + B + C = 5
If each person must receive 1, then we are left with the linear equation:
a + b + c = 2
No of ways to distribute the identical marbles = 4! / (2! * 2!) = 6 ways
(36 ways to distribute the 4 toys) * (6 ways to distribute the 5 identical marbles) = 216 ways
D 216
Hopefully I didn’t miss anything (and read the question correctly)
Posted from my mobile device
I believe the question is saying “at least one toy and at least one marble”
(1st) distributing the 4 different toys to 3 different people
Each person must receive at least 1 of the 4 toys. This means one person will receive 2 toys and the other two people will receive just 1 toy.
[2 ; 1 ; 1]
Step 1: distribute the different toys into identical stacks as listed above
“4 choose 2” * “2 choose 1” * 1 choose 1” = 4! / (2! * 1! * 1!)
However, since there are identical amounts of toys placed in the stacks that contain 1, we will have repeat counts. To remove the overcounting, we need to divide by 2!
End up with: 4! / (2! * 2!*) = 6 ways
AND
Step 2: Because we have 3 distinct groups of people, we need to arrange the “stacks” among the 3 people. Because each stack has different items, it matters which stack with 1 toy goes to which person.
We can arrange the 3 different stacks among 3 different people in 3! Ways
Toys can be distributed in: 6 * 3! = 36 ways
And
(2nd) distribute 5 identical marbles among 3 distinct “groups” (people)
Letting the 3 different people be variables A B and C
A + B + C = 5
If each person must receive 1, then we are left with the linear equation:
a + b + c = 2
No of ways to distribute the identical marbles = 4! / (2! * 2!) = 6 ways
(36 ways to distribute the 4 toys) * (6 ways to distribute the 5 identical marbles) = 216 ways
D 216
Hopefully I didn’t miss anything (and read the question correctly)
Posted from my mobile device
17 Feb 2021, 18:23
Kudos
Bookmarks
The 4 toys can be distributed in 2 , 1, 1 ways where each gets at least 1 toy.
The number of ways of doing this is 4C2 * 2C1 * 1C1 = (4 * 3/2) * 2 * 1 = 12 ways
Now the number of ways where 2,1,1 can be rearranged = 3!/2! = 3 ways (211) (121)(112)
Total ways of distributing the toys = 12 * 3 = 36 ways
For the marbles, since they are the same, let us give away 1 marble to each. We are left with 2 marbles. The number of ways of distributing them is
2, 0, 0 = 3!/2! = 3 ways and
1, 1, 0 = 3!/2! = 3 ways
The total ways = 3 + 3 = 6
Therefore the total distribution can be done in 36 * 6 = 216 ways
Option D
Arun Kumar
The number of ways of doing this is 4C2 * 2C1 * 1C1 = (4 * 3/2) * 2 * 1 = 12 ways
Now the number of ways where 2,1,1 can be rearranged = 3!/2! = 3 ways (211) (121)(112)
Total ways of distributing the toys = 12 * 3 = 36 ways
For the marbles, since they are the same, let us give away 1 marble to each. We are left with 2 marbles. The number of ways of distributing them is
2, 0, 0 = 3!/2! = 3 ways and
1, 1, 0 = 3!/2! = 3 ways
The total ways = 3 + 3 = 6
Therefore the total distribution can be done in 36 * 6 = 216 ways
Option D
Arun Kumar
