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Intern  Joined: 07 Aug 2010
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In how many ways 5 different balls can be arranged in to 3  [#permalink]

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Question Stats: 17% (00:28) correct 83% (01:05) wrong based on 51 sessions

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In how many ways 5 different balls can be arranged in to 3 different boxes so that no box remains empty?
Math Expert V
Joined: 02 Sep 2009
Posts: 59020
Re: how to solve these ps question  [#permalink]

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7
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.
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Re: In how many ways 5 different balls can be arranged in to 3  [#permalink]

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5
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

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GMAT 1: 750 Q50 V40 Re: how to solve these ps question  [#permalink]

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1
2
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
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Manager  Joined: 05 Nov 2012
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Re: how to solve these ps question  [#permalink]

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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
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Joined: 17 Dec 2012
Posts: 623
Location: India
Re: how to solve these ps question  [#permalink]

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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 $$3 ^ 5$$
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Manager  Joined: 05 Nov 2012
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Re: In how many ways 5 different balls can be arranged in to 3  [#permalink]

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thank you Srinivasan
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Posts: 9
Re: In how many ways 5 different balls can be arranged in to 3  [#permalink]

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1
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...

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

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Bunuel wrote:
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.

Hey, doesn't the (n-1)C(r-1) formula apply here?
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Re: In how many ways 5 different balls can be arranged in to 3  [#permalink]

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mbaiseasy wrote:
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

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

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Bunuel wrote:
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.

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?
Math Expert V
Joined: 02 Aug 2009
Posts: 8157
Re: In how many ways 5 different balls can be arranged in to 3  [#permalink]

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vanam52923 wrote:
Bunuel wrote:
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.

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?

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

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Bunuel wrote:
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.

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! Re: In how many ways 5 different balls can be arranged in to 3   [#permalink] 29 Oct 2019, 08:59
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