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

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Five balls of different colors are to be placed in three boxes of diff [#permalink]
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chetan2u wrote:
Bunuel wrote:
Five balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we place the balls so that no box remains empty?

A. 75
B. 90
C. 120
D. 150
E. 180

We have 5 different balls and 3 different boxes. There is at least one ball in each box, so the different combinations become very limited and can be worked out individually.

1) 3-1-1
This means that one of the box has 3 balls, while the other two have one each
Ways to choose Balls : we can choose 3 balls out of 5 in 5C3 ways, and then choose 1 out of remaining 2 in 2C1 ways => 5C3*2C1*1C1=20 ways
Ways to choose boxes: we have 3 in one and two of them have similar number, that is 1. The permutations when 2 are similar are 3!/2!=3 ways.
Total ways = 20*3=60 ways

2) 2-2-1......The explanation is similar to the first case.
Balls in 5C2*3C2=30
boxes in 3!/2!=3 ways.
Total 30*3=90

Total 60+90=150

D

Hi chetan2u ,
In one of such at least questions, you use the approach --- first seat one ball in each basket and then use n-1 C r-1 .
Why have you not used that approach in this question ?

Originally posted by ShankSouljaBoi on 10 Feb 2021, 11:50.
Last edited by ShankSouljaBoi on 10 Feb 2021, 21:03, edited 1 time in total.
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Re: Five balls of different colors are to be placed in three boxes of diff [#permalink]
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chetan2u wrote:
Bunuel wrote:
Five balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we place the balls so that no box remains empty?

A. 75
B. 90
C. 120
D. 150
E. 180

We can calculate each way.
1) 3-1-1
Balls in 5C3*2C1=20 and boxes in 3!/2!=3 ways. Total ways = 20*3=60 ways
2) 2-2-1
Balls in 5C2*3C2=30 and boxes in 3!/2!=3 ways. Total 30*3=90

Total 60+90=150

D

can you explsin the meaning as well, please?

Posted from my mobile device

Please check the solution now. I have added the details. If any further query, please let me know.
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Five balls of different colors are to be placed in three boxes of diff [#permalink]
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ShankSouljaBoi wrote:
chetan2u wrote:
Bunuel wrote:
Five balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we place the balls so that no box remains empty?

A. 75
B. 90
C. 120
D. 150
E. 180

We have 5 different balls and 3 different boxes. There is at least one ball in each box, so the different combinations become very limited and can be worked out individually.

1) 3-1-1
This means that one of the box has 3 balls, while the other two have one each
Ways to choose Balls : we can choose 3 balls out of 5 in 5C3 ways, and then choose 1 out of remaining 2 in 2C1 ways => 5C3*2C1*1C1=20 ways
Ways to choose boxes: we have 3 in one and two of them have similar number, that is 1. The permutations when 2 are similar are 3!/2!=3 ways.
Total ways = 20*3=60 ways

2) 2-2-1......The explanation is similar to the first case.
Balls in 5C2*3C2=30
boxes in 3!/2!=3 ways.
Total 30*3=90

Total 60+90=150

D

Hi chetan2u ,
In one of such at least questions, you use the approach --- first seat one ball in each basket and then use n+r-1 C r-1 .
Why have you not used that approach in this question ?

Hi,
I am sure that question would be to distribute n identical objects in x boxes or something like that.
But here we have 5 different objects in 3 different boxes, so the formula used in that question cannot be used here.
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Re: Five balls of different colors are to be placed in three boxes of diff [#permalink]
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chetan2u ,

Absolutely right sir, and then you used n-1 C r-1 and not n+r-1 C r-1
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Re: Five balls of different colors are to be placed in three boxes of diff [#permalink]
beautiful conceptual problem. involves the following things:
1. identification of cases - 1,1,3 and 1,2,2
2.Step wise approach

1,1,3
Ways to select balls - 5C1 * 4C1 * 3C3 = 20
Ways to select boxed (BECAUSE they are different) = 3!/2! = 3
20*3 = 60

1,2,2
Ways to select balls - 5C1 * 4C2 * 2C2 = 30
Ways to select boxed (BECAUSE they are different) - 3!/2! = 3
30*3 = 90

Total = 150 (D)
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Re: Five balls of different colors are to be placed in three boxes of diff [#permalink]
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