It is currently 23 Sep 2017, 06:16

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Five balls of different colors are to be placed in three

Author Message
TAGS:

### Hide Tags

Intern
Joined: 29 Jan 2013
Posts: 44

Kudos [?]: 45 [0], given: 24

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

15 Nov 2013, 19:38
when can we use (n-1)C(r-1) ?
It is equal to number of ways of dividing n things tor people where each can receive atleast 1? isnt it the case here?

Kudos [?]: 45 [0], given: 24

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7615

Kudos [?]: 16936 [0], given: 230

Location: Pune, India
Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

15 Nov 2013, 23:52
when can we use (n-1)C(r-1) ?
It is equal to number of ways of dividing n things tor people where each can receive atleast 1? isnt it the case here?

The n things need to be identical to use this formula.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 16936 [0], given: 230

Intern
Joined: 29 Jan 2013
Posts: 44

Kudos [?]: 45 [0], given: 24

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

16 Nov 2013, 01:05
oh got it...thanks for pointing out the difference..!!

Kudos [?]: 45 [0], given: 24

Manager
Joined: 18 Jul 2013
Posts: 72

Kudos [?]: 93 [0], given: 114

Location: Italy
GMAT 1: 600 Q42 V31
GMAT 2: 700 Q48 V38
Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

17 Nov 2013, 07:10
VeritasPrepKarishma wrote:

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

thank you so much for answering me.

Kudos [?]: 93 [0], given: 114

Intern
Joined: 25 Aug 2013
Posts: 4

Kudos [?]: [0], given: 15

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

25 Nov 2013, 08:13
For at least 1 ball in each box, 2 kind of arrangements possible: (1) 3-1-1 or (2) 2-2-1

(1) First choose balls for this arrangement using combinations formula: 5C3 x 2C1 x 1C1; then Permute the arrangement: 3! / 2!; so we get: 5C3 x 2C1 x 1C1 x 3! / 2!=60;

(2) Similarly: 5C2 x 3C2 x 1C1 x 3! / 2!=90;

do (1)+(2)=60+90=150

Kudos [?]: [0], given: 15

Intern
Joined: 19 Mar 2013
Posts: 23

Kudos [?]: 4 [0], given: 27

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

12 Dec 2013, 23:04
We have to ways of putting ball in the boxes

1) 2+2+1 which gives (5!3/2!2!1!)=90 ways, where 5! is number of ways to arrange 5 balls in a row, multiply by 3 as we have 3 boxes, divide by 2!2!1! to un-arrange balls in the boxes as the order doesn't matter
2) 3+1+1 which gives 5!3/3!1!1!=60 ways, the same reasoning as in case 1
3) 90+60=150 adding, as either case works for us

Kudos [?]: 4 [0], given: 27

Moderator
Joined: 20 Dec 2013
Posts: 185

Kudos [?]: 74 [1], given: 71

Location: United States (NY)
GMAT 1: 640 Q44 V34
GMAT 2: 710 Q48 V40
GMAT 3: 720 Q49 V40
GPA: 3.16
WE: Consulting (Venture Capital)
Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

02 Feb 2014, 00:46
1
KUDOS
Leveraging off of Bunuel's approach, but attempting to tackle from a slightly different angle.

3-1-1:

Ball Combos-
(Choose 3 from 5)*(Choose 1 from 2 remaining)*(Choose 1 from 1 remaining)
5C3 * 2C1 * 1C1 = 20

Grouping Arrangements-
(Choose 2 "1 groupings" from 3 slots with the 3 grouping automatically filling the vacant spot)
3C2 = 3

20*3=60

2-2-1:

Ball Combos-
(Choose 2 from 5)*(Choose 2 from 3 remaining)*(Choose 1 from 1 remaining)
5C2 * 3C2 *1C1 = 30

Grouping Arrangements-
(Choose 2 "2 groupings" from 3 slots, with the 1 grouping automatically filling the vacant spot)
3C2 = 3

30*3=90

60+90=150
_________________

Kudos [?]: 74 [1], given: 71

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 17615

Kudos [?]: 271 [0], given: 0

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

07 Mar 2015, 12:23
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 271 [0], given: 0

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 17615

Kudos [?]: 271 [0], given: 0

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

21 Jun 2016, 07:22
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 271 [0], given: 0

Intern
Joined: 05 Apr 2016
Posts: 37

Kudos [?]: 1 [0], given: 23

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

30 Jun 2016, 08:54
Bunuel 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.

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.

[b]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).[/b]

Total: 60+90=150.

Hope it's clear.

On stage 2 (1-2-2) shouldn't it be 3*5C1*2*4C2 ? My reasoning is that for choosing which of the 3 boxes will have 1 ball we have to multiply by 3. Then 5C1 is for which of the 5 balls will go in that box. Then there are 2 boxes left that will contain 2 balls, hence we multiply by 2. Then the remaining balls could go in 4C2 ways to one of those boxes.

What am I doing wrong?

Kudos [?]: 1 [0], given: 23

Math Expert
Joined: 02 Sep 2009
Posts: 41695

Kudos [?]: 124575 [1], given: 12079

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

30 Jun 2016, 09:06
1
KUDOS
Expert's post
Mariwa wrote:
Bunuel 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.

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.

[b]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).[/b]

Total: 60+90=150.

Hope it's clear.

On stage 2 (1-2-2) shouldn't it be 3*5C1*2*4C2 ? My reasoning is that for choosing which of the 3 boxes will have 1 ball we have to multiply by 3. Then 5C1 is for which of the 5 balls will go in that box. Then there are 2 boxes left that will contain 2 balls, hence we multiply by 2. Then the remaining balls could go in 4C2 ways to one of those boxes.

What am I doing wrong?

When you choose with $$C^2_4$$ balls for the second box, the remaining 2 automatically go to the third box, so non need to multiply by 2 here.
_________________

Kudos [?]: 124575 [1], given: 12079

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 17615

Kudos [?]: 271 [0], given: 0

Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

10 Jul 2017, 02:08
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 271 [0], given: 0

Senior Manager
Joined: 13 Mar 2017
Posts: 458

Kudos [?]: 94 [0], given: 53

Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

10 Jul 2017, 02:58
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 5 balls are to be arranged in 3 different boxes and each box should contain at least 1 ball.
This can be done as 2 cases
Case 1 : 3-1-1 i.e. 3 balls in 1 box, 1 ball in 2nd box & 1 ball in 3rd box.
To count the number of different ways for this case, we will go step by step.
1. Select 3 balls from 5 balls which will be in 1 box. = 5C3
2. Naturally the other 2 balls will go in the left 2 boxes.
3. Now this ball arrangement can itself be arranged in 3! ways.
4. Total ways = 5C3 * 3! = 5*4/2/1 * 3*2*1 = 60 ways.

Case 2 : 2-2-1 i.e. 2 balls in 1 box, 2 balls in 2nd box & 1 ball in 3rd box.
Again we will go step by step for this case.
1. Select 2 balls from 5 balls which will be in 1 box. = 5C2
2. Select 2 balls from 3 left balls which will be in 2 box. = 3C2
3. The left 1 ball will be placed in 3rd box.
Total such selection = 5C2*3C2*1 /2 (since 2 ball selected in 1st and 2 ball selected in 2nd will be same in different conditions.)
4. Now this ball arrangement can itself be arranged in 3! ways.
5. total ways = (5C2 * 3C2 * 1/2) * 3!
= 90

Total ways = 60 +90 = 150 ways,
_________________

MBA Social Network : WebMaggu

Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)

Kudos [?]: 94 [0], given: 53

Director
Joined: 12 Dec 2016
Posts: 850

Kudos [?]: 7 [0], given: 798

Location: United States
GMAT 1: 700 Q49 V33
GPA: 3.64
Re: Five balls of different colors are to be placed in three [#permalink]

### Show Tags

28 Aug 2017, 01:48
this is a common pattern in gmat, but test takers should only know this and learn the correct answer, not the solution.

Kudos [?]: 7 [0], given: 798

Re: Five balls of different colors are to be placed in three   [#permalink] 28 Aug 2017, 01:48

Go to page   Previous    1   2   [ 34 posts ]

Similar topics Replies Last post
Similar
Topics:
7 In how many different ways can all of 5 identical balls be placed 2 03 Feb 2017, 03:56
1 In a jar there are balls in different colors: blue, red, green and yel 3 13 Jul 2016, 08:11
9 In a jar there are balls in different colors: blue, red, green and yel 12 09 Oct 2016, 07:49
21 How many ways can 5 different colored marbles be placed in 3 7 18 Aug 2016, 23:47
3 A woman has three blouses of different colors, three skirts of differe 1 17 Jan 2016, 21:11
Display posts from previous: Sort by