It is currently 18 Dec 2017, 01:07

### 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

# In how many ways 5 different balls can be arranged in to 3

Author Message
TAGS:

### Hide Tags

Intern
Joined: 07 Aug 2010
Posts: 1

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

In how many ways 5 different balls can be arranged in to 3 [#permalink]

### Show Tags

07 Aug 2010, 05:10
7
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

29% (00:28) correct 71% (00:55) wrong based on 23 sessions

### HideShow timer Statistics

In how many ways 5 different balls can be arranged in to 3 different boxes so that no box remains empty?

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

Math Expert
Joined: 02 Sep 2009
Posts: 42669

Kudos [?]: 136009 [2], given: 12723

Re: how to solve these ps question [#permalink]

### Show Tags

08 Aug 2010, 04:27
2
KUDOS
Expert's post
3
This post was
BOOKMARKED
aaratigarlapati wrote:
In how many ways 5 different balls can be arranged in to 3 different boxes so that no box remains empty?

Boxes: I - II - III
Balls: a, b, c, d, e.

There are only 2 distributions possible 3-1-1 or 1-2-2.

3-1-1 case: $$C^1_3*C^3_5*2=60$$.
$$C^1_3$$ - # of ways to choose which box will get 3 balls;
$$C^3_5$$ - # of ways to choose which 3 balls will get this chosen box;
$$2$$ - # of ways to place 2 balls left in 2 other boxes.

1-2-2 case: $$C^1_3*C^1_5*C^2_4=90$$.
$$C^1_3$$ - # of ways to choose which box will get 1 ball;
$$C^1_5$$ - # of ways to choose which ball will get this chosen box;
$$C^2_4$$ - # of ways to choose which 2 balls will get the second box (the third box will get the left 2 balls).

Total: 60+90=150.
_________________

Kudos [?]: 136009 [2], given: 12723

CEO
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4724 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: how to solve these ps question [#permalink]

### Show Tags

08 Aug 2010, 05:45
Expert's post
1
This post was
BOOKMARKED
another way:

1. The total number of possibilities including empty boxes: $$3^5 = 243$$

2. Two of the boxes are empty: $$C^3_2= 3$$

3. One but not two of the boxes is empty: $$3*(2^5 - 2) = 90$$

4. the total number of possibilities excluding empty boxes: 243 - 3 - 90 = 150
_________________

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

Kudos [?]: 4724 [0], given: 360

Manager
Joined: 05 Nov 2012
Posts: 163

Kudos [?]: 40 [0], given: 57

Re: how to solve these ps question [#permalink]

### Show Tags

24 Dec 2012, 09:50
walker wrote:
another way:
3. One but not two of the boxes is empty: $$3*(2^5 - 2) = 90$$

Can you explain this part..... thanks

Kudos [?]: 40 [0], given: 57

Director
Joined: 17 Dec 2012
Posts: 623

Kudos [?]: 550 [0], given: 16

Location: India
Re: how to solve these ps question [#permalink]

### Show Tags

25 Dec 2012, 06:37
Amateur wrote:
walker wrote:
another way:
3. One but not two of the boxes is empty: $$3*(2^5 - 2) = 90$$

Can you explain this part..... thanks

Assume one box is empty. Now take the other 2 boxes out of 3 boxes. The number of ways of placing the 5 balls in 2 boxes is $$2 ^ 5$$ . But this includes the combinations where 5 balls placed in one of them and the other is empty. The number of such combinations is 2 as we are considering 2 boxes. So subtract that from $$2^ 5$$ as we are considering only one empty box, which we assumed at the beginning. That gives us$$2^5 - 2$$. Now we have actually 3 boxes. So any 2 boxes as above can be selected out of the 3 boxes in 3 ways. Therefore the we have$$3 * (2^5 - 2)$$

The above gives the number of combinations where only 1 box is empty. This has to be added to the number of possibilities of two boxes being empty which is 3. The total is 93. This has to be subtracted from the total number of possibilities of $$3 ^ 5$$
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com/regularcourse.php

Standardized Approaches

Kudos [?]: 550 [0], given: 16

Manager
Joined: 05 Nov 2012
Posts: 163

Kudos [?]: 40 [0], given: 57

Re: In how many ways 5 different balls can be arranged in to 3 [#permalink]

### Show Tags

25 Dec 2012, 07:12
thank you Srinivasan

Kudos [?]: 40 [0], given: 57

Senior Manager
Joined: 13 Aug 2012
Posts: 457

Kudos [?]: 573 [3], given: 11

Concentration: Marketing, Finance
GPA: 3.23
Re: In how many ways 5 different balls can be arranged in to 3 [#permalink]

### Show Tags

27 Dec 2012, 22:37
3
KUDOS
4
This post was
BOOKMARKED
aaratigarlapati wrote:
In how many ways 5 different balls can be arranged in to 3 different boxes so that no box remains empty?

How can we feel three boxes with 5 balls?
3 + 1 + 1
2 + 2 + 1

With 3,1,1 distribution:
How many ways to select 3 from 5?
5!/3!2! = 10
How many ways to select 1 ball from 2?
2!/1! = 2
How many ways to select 1 ball from 1?
1!/1! = 1
How many ways to distribute 3,1 and 1 to 3 boxes? 3!/2! = 3

10*2*3 = 60

With 2,2,1 distribution:
How many ways to select 2 from 5?
5!/2!3! = 10
How many ways to select 2 from 3?
3!/2!1! = 3
How many ways to select 1 from 1?
1
How many ways to distribute 2,2,1 to 3 boxes? 3!/2! = 3

10*3*3=90

90+60 = 150

_________________

Impossible is nothing to God.

Kudos [?]: 573 [3], given: 11

Non-Human User
Joined: 09 Sep 2013
Posts: 14780

Kudos [?]: 288 [0], given: 0

Re: In how many ways 5 different balls can be arranged in to 3 [#permalink]

### Show Tags

18 Dec 2015, 11:43
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 [?]: 288 [0], given: 0

Non-Human User
Joined: 09 Sep 2013
Posts: 14780

Kudos [?]: 288 [0], given: 0

Re: In how many ways 5 different balls can be arranged in to 3 [#permalink]

### Show Tags

11 Sep 2017, 00:19
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 [?]: 288 [0], given: 0

Manager
Status: Preparing
Joined: 05 May 2016
Posts: 62

Kudos [?]: 12 [0], given: 150

Location: India
Re: In how many ways 5 different balls can be arranged in to 3 [#permalink]

### Show Tags

27 Sep 2017, 08:15
aaratigarlapati wrote:
In how many ways 5 different balls can be arranged in to 3 different boxes so that no box remains empty?

Can someone please explain the solution once again?I have gone through the above solutions but not able to understand those.

Kudos [?]: 12 [0], given: 150

Re: In how many ways 5 different balls can be arranged in to 3   [#permalink] 27 Sep 2017, 08:15
Display posts from previous: Sort by