Find all School-related info fast with the new School-Specific MBA Forum

It is currently 21 Oct 2014, 06:22

Close

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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Five balls of different colors are to be placed in three

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 11 Jul 2012
Posts: 52
GMAT 1: 650 Q49 V29
Followers: 0

Kudos [?]: 6 [0], given: 20

Five balls of different colors are to be placed in three [#permalink] New post 18 Oct 2012, 13:12
10
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

35% (02:57) correct 65% (01:12) wrong based on 118 sessions
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


Please provide a small note of explanation for all the combinations used in the solution.
[Reveal] Spoiler: OA
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23351
Followers: 3603

Kudos [?]: 28706 [1] , given: 2811

Re: Five balls of different colors are to be placed in three [#permalink] New post 18 Oct 2012, 13:29
1
This post received
KUDOS
Expert's post
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


We can have the following two distributions:

1. 3-1-1 one box gets three balls and the remaining two boxes get one ball each.

The number of ways to distribute the balls for this case is 3*C^3_5*2=60, where 3 is the number of ways to choose which box gets 3 balls (we have 3 boxes, thus 3 choices for that), C^3_5 is the number of ways to choose which 3 balls out of 5 will go to that box, and 2 is the number of ways to distribute the remaining 2 balls in the remaining two boxes.

2. 1-2-2 one box gets one ball and the remaining two boxes get two balls each.

The number of ways to distribute the balls for this case is 3*5*C^2_4=90, where 3 is the number of ways to choose which box gets 1 balls (we have 3 boxes, thus 3 choices for that), 5 is the number of ways to choose which ball out of 5 will go to that box, and C^2_4 is the number of ways to chose which 2 balls out of 4 balls left will go to the second box (the remaining 2 balls will naturally go to the third box).

Total: 60+90=150.

Answer: D.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

4 KUDOS received
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 73

Kudos [?]: 529 [4] , given: 43

Re: Five balls of different colors are to be placed in three [#permalink] New post 18 Oct 2012, 13:49
4
This post received
KUDOS
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box.
For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box.
Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes.
For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls.
For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities.

Total of 90 + 60 = 150 possibilities.

Answer D.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Expert Post
CEO
CEO
User avatar
Joined: 17 Nov 2007
Posts: 3571
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 367

Kudos [?]: 1849 [0], given: 358

GMAT ToolKit User Premium Member
Re: Five balls of different colors are to be placed in three [#permalink] New post 18 Oct 2012, 14:18
Expert's post
Another way:

N = 3^5 - 3*2^5 + 3 = 243 - 96 + 3 = 150

3^5 - the total number of ways to put 5 different balls into 3 different boxes. Each ball has 3 possible options (boxes)

3*2^5 - 3 - the total number of ways to put 5 different balls into 1 or 2 different boxes out of 3 (to leave at least 1 box empty).
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Manager
avatar
Joined: 11 Jul 2012
Posts: 52
GMAT 1: 650 Q49 V29
Followers: 0

Kudos [?]: 6 [0], given: 20

Re: Five balls of different colors are to be placed in three [#permalink] New post 18 Oct 2012, 19:03
EvaJager wrote:
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box.
For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box.
Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes.
For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls.
For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities.

Total of 90 + 60 = 150 possibilities.

Answer D.


Could you please break down the highlighted part and explain how are you getting those figures.....
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 73

Kudos [?]: 529 [0], given: 43

Re: Five balls of different colors are to be placed in three [#permalink] New post 19 Oct 2012, 06:07
avaneeshvyas wrote:
EvaJager wrote:
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box.
For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box.
Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes.
For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls.
For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities.

Total of 90 + 60 = 150 possibilities.

Answer D.


Could you please break down the highlighted part and explain how are you getting those figures.....


If there would not have been boxes, 5 different balls we could arrange in 5! = 120 ways.
Now look at a given arrangement of the 5 balls, let's say ABCDE. Because we have three different boxes, then the following arrangements are possible:

|AB| |CD| |E|; |AB| |C| |DE|; |A| |BC| |DE| - these are all 2,2,1 type scenarios

and they are considered different arrangements (this would give 5! * 3 possibilities), EXCEPT THAT, inside each box, if we have more than one ball, order doesn't matter. So, when in a box we have for example AB, this is the same as BA, we don't mind whether A or B was put first into the box. Therefore, we should divide by k! whenever we have in a box k balls. 1! doesn't change the value of the expression, I left them just for the completeness of the formula.

Similar reasoning applies for the 3,1,1 scenario.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Manager
avatar
Joined: 16 Feb 2011
Posts: 197
Schools: ABCD
Followers: 1

Kudos [?]: 40 [0], given: 78

Re: Five balls of different colors are to be placed in three [#permalink] New post 19 Oct 2012, 18:21
EvaJager wrote:
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box.
For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box.
Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes.
For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls.
For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities.

Total of 90 + 60 = 150 possibilities.

Answer D.



Eva,
I was able to compute the combinations for distributing balls in 2-2-1 and 3-1-1 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z -- and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one?

Example : abcde are balls and three boxes are xyz =>
X=[AB] Y=[cd] Z=[e]
X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6

I am a bit confused. Still thinking....Please help me. thanks
1 KUDOS received
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 73

Kudos [?]: 529 [1] , given: 43

Re: Five balls of different colors are to be placed in three [#permalink] New post 19 Oct 2012, 23:10
1
This post received
KUDOS
voodoochild wrote:
EvaJager wrote:
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box.
For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box.
Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes.
For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls.
For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities.

Total of 90 + 60 = 150 possibilities.

Answer D.



Eva,
I was able to compute the combinations for distributing balls in 2-2-1 and 3-1-1 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z -- and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one?

Example : abcde are balls and three boxes are xyz =>
X=[AB] Y=[cd] Z=[e]
X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6

I am a bit confused. Still thinking....Please help me. thanks


You keep the boxes fixed, you permute the balls.
For example, if you have in the boxes |AB| |CD| |E|. This is obtained from ABCDE.
You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls.

For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities:
|AB| |CD| |E|
|AB| |C| |DE|
|A| |BC| |DE|

So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example:
AB in the first box is the same as BA, and CD in the second box is the same as DC.
All the arrangements
|AB| |CD| |E|
|BA| |CD| |E|
|AB| |DC| |E|
|BA| |DC| |E|
should be counted just once.

Therefore, 5!*3/(2! * 2! * 1!).
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Expert Post
3 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4877
Location: Pune, India
Followers: 1151

Kudos [?]: 5348 [3] , given: 165

Re: Five balls of different colors are to be placed in three [#permalink] New post 20 Oct 2012, 04:13
3
This post received
KUDOS
Expert's post
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts.

In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty)
Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball.
Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways

Only one box empty:
In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty)
Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5 - 2).

2 boxes empty:
In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways.

You can distribute 5 balls in 3 boxes such that no box is empty in 3^5 - 3*(2^5 - 2) - 3 = 243 - 90 - 3 = 150 ways

For a discussion on distribution of objects, check out these posts too:
http://www.veritasprep.com/blog/2011/12 ... 93-part-1/
http://www.veritasprep.com/blog/2011/12 ... s-part-ii/
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

1 KUDOS received
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 73

Kudos [?]: 529 [1] , given: 43

Re: Five balls of different colors are to be placed in three [#permalink] New post 20 Oct 2012, 11:26
1
This post received
KUDOS
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


This is the beauty of combinatorics: doesn't matter how you execute your plan (here, that of distributing some balls in boxes), if the stages are correctly translated into mathematical expressions, you always get the correct answer. We have already seen three different approaches to solve the question.
Just for the fun, I tried to reach the correct answer with another scenario:

Two stages:
1) Distribute one ball in each box, to be sure that there is at least one ball in each box. This can be done in 5*4*3 = 60 ways, boxes are distinct, order matters.
Continue with stage:
2) Place the remaining 2 balls.
Here we have two possibilities:
2a) Either both balls will be placed in one of the 3 boxes - OR -
2b) Each ball will be placed in a separate box.

For 1 and 2a) - 60*3/3 = 60; 3 possibilities to choose the box that will get the remaining 2 balls; divide by 3, because in the box with the 3 balls, it doesn't matter which ball was placed in at stage 1, it could be any of the final 3 balls.

For 1 and 2b) - 60*3*2/4 = 90; 3 possibilities to choose the box that will not get an extra ball, which is equivalent to choosing 2 boxes that will each get an extra ball; 2 possibilities to place the remaining balls in the 2 different boxes; divide by 4 = 2*2, because in those 2 boxes with 2 balls, it doesn't matter in which order the 2 balls arrived, stage 1 or stage 2b).

Again, we obtain the total of 60 + 90 = 150 possibilities.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Manager
avatar
Joined: 16 Feb 2011
Posts: 197
Schools: ABCD
Followers: 1

Kudos [?]: 40 [0], given: 78

Re: Five balls of different colors are to be placed in three [#permalink] New post 20 Oct 2012, 18:19
EvaJager wrote:
voodoochild wrote:
EvaJager wrote:
Since the balls are all of different colors, let's permute them and then decide how many balls we put in each box.
For example, arrange in a row the balls, then decide: two balls go into the first box, next two in the second box, and the last ball goes to the third box.
Since in each box there must be at least one ball, we have the possibilities of (2,2,1), (2,1,2), (1,2,2) OR (3,1,1), (1,3,1), (1,1,3) balls in the three boxes.
For the 2,2,1 type arrangements, we have [5!/(2!2!1!)]*3 = 90 possibilities. Inside a box, it doesn't matter the order of the balls.
For the 3,1,1 type arrangements, we have [5!/(3!1!1!)]*3 = 60 possibilities.

Total of 90 + 60 = 150 possibilities.

Answer D.



Eva,
I was able to compute the combinations for distributing balls in 2-2-1 and 3-1-1 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z -- and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one?

Example : abcde are balls and three boxes are xyz =>
X=[AB] Y=[cd] Z=[e]
X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6

I am a bit confused. Still thinking....Please help me. thanks


You keep the boxes fixed, you permute the balls.
For example, if you have in the boxes |AB| |CD| |E|. This is obtained from ABCDE.
You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls.

For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities:
|AB| |CD| |E|
|AB| |C| |DE|
|A| |BC| |DE|

So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example:
AB in the first box is the same as BA, and CD in the second box is the same as DC.
All the arrangements
|AB| |CD| |E|
|BA| |CD| |E|
|AB| |DC| |E|
|BA| |DC| |E|
should be counted just once.

Therefore, 5!*3/(2! * 2! * 1!).


Eva,
Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will matter -correct?), then we will have to multiply by 3!? That is, for 2-2-1 case, 5C2*3C2*1*(3!) ---Please let me know your thoughts....

{ My approach:
Here's how I approached this problem :- for 2-2-1 case, the combinations will be 5C2*3C2*1C1 (choose 2 balls from 5 balls - then - choose 2 balls from the remaining 3 balls - then - choose 1 ball from 1 ball)

Now, since the boxes are labeled differently, we need to permute them. I initially thought of multiplying by 3! because of these combinations :
X[AB] Y[cd] Z[e]
X[AB] Z[e] Y[cd]
X[CD] Y[AB] Z[e]
X[CD] Z[e] Y[AB]
X[e] Y[AB] Z[CD]
X[e] Z[CD] Y[AB]
........ditto for 3-1-1 case.
}

I am a bit confused.


Thanks
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 73

Kudos [?]: 529 [0], given: 43

Re: Five balls of different colors are to be placed in three [#permalink] New post 21 Oct 2012, 03:18
voodoochild wrote:
EvaJager wrote:
voodoochild wrote:

Eva,
I was able to compute the combinations for distributing balls in 2-2-1 and 3-1-1 way. However, I used 3! for the arrangement of the boxes. Let's say that the boxes are labeled X Y and Z -- and we are using (2,2,1) type of placement of balls. Wouldn't the order of boxes matter? i.e. 3!? That is, I could have any of the three boxes in the first spot, 2 for the second spot and 1 for the third one?

Example : abcde are balls and three boxes are xyz =>
X=[AB] Y=[cd] Z=[e]
X[AB] Z[e] Y[cd] .... (repeat two rows three times) = 6

I am a bit confused. Still thinking....Please help me. thanks


You keep the boxes fixed, you permute the balls.
For example, if you have in the boxes |AB| |CD| |E|. This is obtained from ABCDE.
You also have |CD| |E| |AB|, which is obtained from CDEAB, an arrangement already included in the 5! permutations of the balls.

For any permutation of the balls, in the 2,2,1, scenario (meaning there are two boxes with two balls and one box with one ball), you need the factor of 3 because there are 3 possibilities:
|AB| |CD| |E|
|AB| |C| |DE|
|A| |BC| |DE|

So, you have a total of 5! * 3 arrangements. But here we count each arrangement 4 times, which is 2! * 2!. For example:
AB in the first box is the same as BA, and CD in the second box is the same as DC.
All the arrangements
|AB| |CD| |E|
|BA| |CD| |E|
|AB| |DC| |E|
|BA| |DC| |E|
should be counted just once.

Therefore, 5!*3/(2! * 2! * 1!).


Eva,
Thanks for your help. I think that I see your point. Do you think that if the boxes are placed on a rack in a store (order of the three boxes will matter -correct?), then we will have to multiply by 3!? That is, for 2-2-1 case, 5C2*3C2*1*(3!) ---Please let me know your thoughts....

{ My approach:
Here's how I approached this problem :- for 2-2-1 case, the combinations will be 5C2*3C2*1C1 (choose 2 balls from 5 balls - then - choose 2 balls from the remaining 3 balls - then - choose 1 ball from 1 ball)

Now, since the boxes are labeled differently, we need to permute them. NO! I initially thought of multiplying by 3! because of these combinations :
X[AB] Y[cd] Z[e]
X[AB] Z[e] Y[cd]
X[CD] Y[AB] Z[e]
X[CD] Z[e] Y[AB]
X[e] Y[AB] Z[CD]
X[e] Z[CD] Y[AB]
........ditto for 3-1-1 case.
}

I am a bit confused.

Thanks



You don't permute the boxes! Once the boxes are defined, they are unique. You called them X, Y, Z. Place them on a shelf, nailed them down, don't touch them!!! The question is not about arranging the boxes on a shelf, it's about their contents. Even if you would arrange the boxes in a different order, say Y, X, Z, what matters is only their content and not the order in which they are displayed.

For the combinatorial approach, see Bunuel and Walker's solutions.

Here, I was playing a different game. I was permuting the balls and then decide (define) the boxes.

Try to understand each solution separately, and don't mix between them.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 613
WE: Science (Education)
Followers: 73

Kudos [?]: 529 [0], given: 43

Re: Five balls of different colors are to be placed in three [#permalink] New post 21 Oct 2012, 04:22
VeritasPrepKarishma wrote:
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts.

In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty)
Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball.
Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways

Only one box empty:
In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty)
Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5 - 2).

2 boxes empty:


In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways.

You can distribute 5 balls in 3 boxes such that no box is empty in 3^5 - 3*(2^5 - 2) - 3 = 243 - 90 - 3 = 150 ways

For a discussion on distribution of objects, check out these posts too:
http://www.veritasprep.com/blog/2011/12 ... 93-part-1/
http://www.veritasprep.com/blog/2011/12 ... s-part-ii/


Just a remark:

...avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small).

There is no need for enumeration, I wrote them down because the list is indeed short. For each scenario, there are 3C1=3, 3C2=3, possibilities, respectively.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Manager
avatar
Joined: 13 Jul 2013
Posts: 77
GMAT 1: 570 Q46 V24
Followers: 0

Kudos [?]: 4 [0], given: 21

Re: Five balls of different colors are to be placed in three [#permalink] New post 29 Aug 2013, 02:09
VeritasPrepKarishma wrote:
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts.

In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty)
Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball.
Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways

Only one box empty:
In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty)
Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5 - 2).


2 boxes empty:
In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways.

You can distribute 5 balls in 3 boxes such that no box is empty in 3^5 - 3*(2^5 - 2) - 3 = 243 - 90 - 3 = 150 ways

For a discussion on distribution of objects, check out these posts too:
http://www.veritasprep.com/blog/2011/12 ... 93-part-1/
http://www.veritasprep.com/blog/2011/12 ... s-part-ii/


Can someone please explain the text I have marked in Red?
Expert Post
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4877
Location: Pune, India
Followers: 1151

Kudos [?]: 5348 [0], given: 165

Re: Five balls of different colors are to be placed in three [#permalink] New post 29 Aug 2013, 02:42
Expert's post
theGame001 wrote:
VeritasPrepKarishma wrote:
avaneeshvyas 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


Please provide a small note of explanation for all the combinations used in the solution.


I did it the way walker did since we can avoid enumerating by using that method i.e. we avoid (3, 1, 1), (2, 2, 1) etc. When required to enumerate, there is a chance that we might forget one of the possible combinations (though in this question it doesn't matter much since number of balls is small). Do visit that solution properly too. In this question, consider it an exercise in lots of PnC concepts.

In how many ways can you distribute 5 different balls in 3 different boxes such that a box may get no ball? (we will later subtract those cases in which one or two boxes are empty)
Each ball can be placed in any of the 3 boxes so there are 3 ways to place a ball.
Since there are 5 balls, we can place all the balls in 3*3*3*3*3 = 3^5 ways

Only one box empty:
In how many ways can you distribute the 5 different balls in 2 different boxes such that both boxes have at least one ball? (we want to find the cases in which one and only one box will be empty)
Select the empty box in 3 ways. Each ball can be placed in any of the 2 boxes. So 5 balls can be placed in 2*2*2*2*2 = 2^5 ways. Out of 2^5, subtract those cases where the balls are all in 1 box only. This happens in 2 ways since you can select an empty box again out of the two in 2 ways. Total number of ways = 3*(2^5 - 2).


2 boxes empty:
In how many ways can you distribute the 5 balls in only 1 box? In 3 ways since you select a box in 3 ways.

You can distribute 5 balls in 3 boxes such that no box is empty in 3^5 - 3*(2^5 - 2) - 3 = 243 - 90 - 3 = 150 ways

For a discussion on distribution of objects, check out these posts too:
http://www.veritasprep.com/blog/2011/12 ... 93-part-1/
http://www.veritasprep.com/blog/2011/12 ... s-part-ii/


Can someone please explain the text I have marked in Red?


You have 3 boxes. YOu want to keep exactly one empty. This means the other two must have at least one ball. Say the boxes are A, B and C. In how many ways can you choose to have exactly one box empty?
A empty, B and C non empty
B empty, A and C non empty
C empty, A and B non empty

Basically you can select the box you will keep empty in 3 ways.

Now you have to put 5 balls in 2 boxes (say e.g. B and C) such that both boxes have at least one ball.

For each ball, there are two options: box B or box C
So you can put in 5 balls in 2*2*2*2*2 = 2^5 ways.
But this includes cases where all ball are put in one box only. How many such cases are there? All balls could go in box B and box C could be empty. All balls could go in box C and box B would be empty. Only these two cases are possible. So out of the 2^5 cases, you need to remove these two cases because you must have at least one ball in each of the two boxes, B and C.

No of ways in which you have all balls in exactly two boxes = 3*(2^5 - 2)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Manager
Manager
avatar
Joined: 12 Feb 2012
Posts: 108
Followers: 1

Kudos [?]: 10 [0], given: 28

How many ways can you distribute 5 marbles in 3 identical ba [#permalink] New post 30 Aug 2013, 12:18
How many ways can you distribute 5 marbles in 3 identical baskets such that each basket gets at least 1 marble.

I am trying to solve this problems using as many approaches:

The correct answer is:
we can pick distribute the marbles in two such ways:
3-1-1 or 2-2-1.
For the 3-1-1. We choose 3 marbles for the 1st basket. (5C3). 1 for the 2nd basket (2C1) and from the remaining marble (1C1).
We then divide by 2!. I know it has something to do with the the fact we are putting 1 marble into the 2 baskets but what is the intuition?? How do you if you have to divide by n!?? This is what I am not getting. After we have (5C3)(2C1)(1C1) does the problem become... " How many ways can we rearrange the digits 311?" Ans: 3!/(2!1!)

If I choose 3 people out of 5 for a soccer game its 5*4*3/3!. I divide by 3! Because order does not matter. How does this translate to the above problem?
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Status: Tomorrow will be a new day...
Joined: 22 Mar 2013
Posts: 943
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)
Followers: 41

Kudos [?]: 408 [0], given: 209

Re: How many ways can you distribute 5 marbles in 3 identical ba [#permalink] New post 30 Aug 2013, 12:40
n-1Cr-1 = 5-1C3-1 = 4C2 = 6

There are direct formula:
1. Distribute n identical objects to r people, such that each can receive 0,1,2,3...n objects.
(n+r-1)C(r-1)

2. Distribute n identical objects to r people, such that each receive at least one object.
(n-1)C(r-1)

This is what I used in above case, identical baskets sounds tricky, but we can place baskets in series 1 2 3.
_________________

Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos :)
My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Senior Manager
Senior Manager
User avatar
Joined: 17 Dec 2012
Posts: 395
Location: India
Followers: 14

Kudos [?]: 192 [0], given: 10

Re: How many ways can you distribute 5 marbles in 3 identical ba [#permalink] New post 30 Aug 2013, 16:42
alphabeta1234 wrote:
How many ways can you distribute 5 marbles in 3 identical baskets such that each basket gets at least 1 marble.

I am trying to solve this problems using as many approaches:

The correct answer is:
we can pick distribute the marbles in two such ways:
3-1-1 or 2-2-1.
For the 3-1-1. We choose 3 marbles for the 1st basket. (5C3). 1 for the 2nd basket (2C1) and from the remaining marble (1C1).
We then divide by 2!. I know it has something to do with the the fact we are putting 1 marble into the 2 baskets but what is the intuition?? How do you if you have to divide by n!?? This is what I am not getting. After we have (5C3)(2C1)(1C1) does the problem become... " How many ways can we rearrange the digits 311?" Ans: 3!/(2!1!)

If I choose 3 people out of 5 for a soccer game its 5*4*3/3!. I divide by 3! Because order does not matter. How does this translate to the above problem?


Hi,

You see the constraint that each basket has to have at least 1 marble. So put 1 marble in each basket. We have 2 marbles remaining. This can be distibuted among 2 baskets or put in just 1 basket. The former can be done in 3C2 ways and the latter in 3C1 ways. So there are totally 6 possible ways you can do this.

I assume the marbles are identical. So you do not have to consider 5C3.
_________________

Srinivasan Vaidyaraman
Sravna Test Prep
http://www.sravna.com

Classroom Courses in Chennai
Online and Correspondence Courses

Manager
Manager
User avatar
Joined: 18 Jul 2013
Posts: 63
Location: Italy
GMAT 1: 600 Q42 V31
Followers: 0

Kudos [?]: 18 [0], given: 111

CAT Tests
Re: Five balls of different colors are to be placed in three [#permalink] New post 14 Nov 2013, 16:51
Hi i did it that way and i don't understand why i'm wrong:

We can have two distributions:

1. 3-1-1 one box gets three balls, the 2 other boxes only 1

It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes)

So 5C3 * 2C1 * 1C1 = 20

2. 1-2-2 one box gets one ball and the remaining two boxes get two balls each.

It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes)

So 5C1 * 4C2 * 2C2 = 30

so 50 solutions.. :evil:
Expert Post
2 KUDOS received
Veritas Prep GMAT Instructor
User avatar
Joined: 16 Oct 2010
Posts: 4877
Location: Pune, India
Followers: 1151

Kudos [?]: 5348 [2] , given: 165

Re: Five balls of different colors are to be placed in three [#permalink] New post 14 Nov 2013, 20:18
2
This post received
KUDOS
Expert's post
oss198 wrote:
Hi i did it that way and i don't understand why i'm wrong:

We can have two distributions:

1. 3-1-1 one box gets three balls, the 2 other boxes only 1

It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes)

So 5C3 * 2C1 * 1C1 = 20

2. 1-2-2 one box gets one ball and the remaining two boxes get two balls each.

It is equivalent as to make 3 groups ordered out of 5 objects (as there are 3 different boxes)

So 5C1 * 4C2 * 2C2 = 30

so 50 solutions.. :evil:


The boxes are different B1, B2, B3
When you split the balls 3-1-1, which box gets 3 balls? You first select a box which gets 3 balls. You can do this in 3 ways - select B1 or B2 or B3.
Now select 3 of the 5 balls for this box in 5C3 ways. Now you have 2 balls and 2 boxes one for each ball so you can distribute them in 2 ways.
This gives 3*5C3*2 = 60.

Same problem for 2-2-1 case. You multiply by another 3 and you get 3*30 = 90

Total 60+90 = 150
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save $100 on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

Re: Five balls of different colors are to be placed in three   [#permalink] 14 Nov 2013, 20:18
    Similar topics Author Replies Last post
Similar
Topics:
8 How many ways can 5 different colored marbles be placed in 3 praffulpatel 2 14 Oct 2013, 07:53
1 There are 6 different colored balls along with their 6 bmwhype2 1 12 Mar 2008, 18:12
Experts publish their posts in the topic In a jar there are balls in different colors: blue, red, ashkrs 3 22 Dec 2007, 12:18
In a jar there are balls in different colors: Blue, Red, alohagirl 4 08 Nov 2007, 02:15
A black sack contains three green balls, five yellow balls, Swagatalakshmi 11 28 Dec 2006, 14:45
Display posts from previous: Sort by

Five balls of different colors are to be placed in three

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page    1   2    Next  [ 27 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.