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

Question

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

A. 100
B. 110
C. 120
D. 130
E. 150

Bunuel · Accepted Answer

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.

mbaiseasy · Answer

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

Answer: 150

walker · Answer

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

Amateur · Answer

Can you explain this part..... thanks

SVaidyaraman · Answer

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

Amateur · Answer

thank you Srinivasan

nidhiprasad · Answer

My solution is :
First satisfy the condition that all the boxes has at least one ball
1st ball has 3 choices
2nd ball has 2 choice 
3rd ball has 1 choice ... here all the boxes have at least one ball.
4th ball has 3 choices( can go to any of the boxes)
5th ball has 3 choices ( can go to any of the boxes)
and all of the boxes can be arranged in 3! ways.

so 3.2.1.3.3.3!=324...

Please help me understand, why this is not the correct way. I know it doesnt match to the above answers .

OhsostudiousMJ · Answer

Hey, doesn't the (n-1)C(r-1) formula apply here?

htmtruong · Answer

How many ways to distribute 3,1 and 1 to 3 boxes? 3!/2! = 3
Can you please help explain, Should there be 3! ways to distribute (3,1,1) to 3 different boxes? Thank you.

vanam52923 · Answer

Bunuel 
 chetan2u 
3-1-1 case :
why did we not multipply it by 3! after  C^1_3*C^1_5*C^2_4=90 .
as the three baloss can be in any one of the bag?

chetan2u · Answer

Hi

3C1 caters for choosing one of the three bags that will have 3 balls. 
So multiplying by 3!, which otherwise would be 3!/2! Or 3 is already taken care by 3C1

celi1210 · Answer

Hi Bunuel,

how likely is it that such a question appears in the gmat exam? After one day of studying this I do get the principle behind this, but it takes me way longer than 2 minutes.

I've read here that "similar items to dissimilar boxes"-questions are more likely to be asked? Thank you!

Regor60 · Answer

Lazy approach.

Each ball has 3 boxes it could be placed in, so

3^5 = 243 arrangements

This includes cases where 1 or 2 boxes are empty, so those cases need to be subtracted.

1 empty leaves 3*2^5 arrangements = 96. This of course also includes cases where 2 are empty.

2 empty leaves 3 choices

243-96-3 = 144

Since we're double counting the 2 empty box case, we know the answer has to be larger than 144.

The only answer that fits is 150

Posted from my mobile device

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

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