Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60
B. 90
C. 120
D. 150
E. 180
Permutation and combination
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 91
- Joined: Fri Jan 17, 2014 7:34 am
- Thanked: 7 times
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marbleparveen110 wrote:Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60
B. 90
C. 120
D. 150
E. 180
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.
Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.
Total ways = 60+90 = 150.
The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
-
- Senior | Next Rank: 100 Posts
- Posts: 91
- Joined: Fri Jan 17, 2014 7:34 am
- Thanked: 7 times
Thank you so much, Mitch. I was thinking on the same lines but was not getting the right answer.GMATGuruNY wrote:Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marbleparveen110 wrote:Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60
B. 90
C. 120
D. 150
E. 180
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.
Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.
Total ways = 60+90 = 150.
The correct answer is D.
Your solutions are always elegant and simple.
Thank you.
-
- Master | Next Rank: 500 Posts
- Posts: 209
- Joined: Thu Jan 12, 2012 12:59 pm
Hey Mitch
Why are you multiplying it by 3?
Number of box options for the box with 3 marbles = 3
Why dont you multiply it by 3!?Because those 3slots can be arranged in 3! ways.Why do you only target the minority always?Let A,B,C be the 3 boxes,
Box A gets 5C3 , Box B gets 2C1 ,Box C gets 1C1 and the boxes can further be rearranged in 3! ways.
We should have a similar case in the 2-2-1 strategy also.
I always find it hard to decide whether to include the factorial or not,Ive done so many problems and have mixed views.Please explain.
Thanks
Why are you multiplying it by 3?
Number of box options for the box with 3 marbles = 3
Why dont you multiply it by 3!?Because those 3slots can be arranged in 3! ways.Why do you only target the minority always?Let A,B,C be the 3 boxes,
Box A gets 5C3 , Box B gets 2C1 ,Box C gets 1C1 and the boxes can further be rearranged in 3! ways.
We should have a similar case in the 2-2-1 strategy also.
I always find it hard to decide whether to include the factorial or not,Ive done so many problems and have mixed views.Please explain.
Thanks
GMATGuruNY wrote:Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marbleparveen110 wrote:Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60
B. 90
C. 120
D. 150
E. 180
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.
Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.
Total ways = 60+90 = 150.
The correct answer is D.
-
- Senior | Next Rank: 100 Posts
- Posts: 91
- Joined: Fri Jan 17, 2014 7:34 am
- Thanked: 7 times
dddanny2006 wrote:Hey Mitch
Why are you multiplying it by 3?
Number of box options for the box with 3 marbles = 3
Why dont you multiply it by 3!?Because those 3slots can be arranged in 3! ways.Why do you only target the minority always?Let A,B,C be the 3 boxes,
Box A gets 5C3 , Box B gets 2C1 ,Box C gets 1C1 and the boxes can further be rearranged in 3! ways.
We should have a similar case in the 2-2-1 strategy also.
I always find it hard to decide whether to include the factorial or not,Ive done so many problems and have mixed views.Please explain.
Thanks
GMATGuruNY wrote:Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marbleparveen110 wrote:Five balls of different colors are to be placed in three different boxes such that any box contains at least 1 ball . What is the maximum number of different ways in which this can be done?
A. 60
B. 90
C. 120
D. 150
E. 180
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.
Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.
Total ways = 60+90 = 150.
The correct answer is D.
You may choose 3 balls out of five in 5C3 ways but these three balls can go into any of the three boxes in three possible ways.
Therefore, 5C3 x 3= 30
now the remaining two balls can go into two different boxes such that both the boxes get one ball each in 2 ways.
Therefore, 30 x 2=60
The same logic is applied in the second case also.