Permutation and combination

parveen110 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

Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marble
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.

GMATGuruNY wrote:

parveen110 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

Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marble
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.

Thank you so much, Mitch. I was thinking on the same lines but was not getting the right answer.
Your solutions are always elegant and simple.
Thank you.

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:

parveen110 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

Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marble
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.

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:

parveen110 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

Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marble
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.