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

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17 Nov 2013, 06:10

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

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

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25 Nov 2013, 07:13

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;

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

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12 Dec 2013, 22:04

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

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

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07 Mar 2015, 11:23

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

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21 Jun 2016, 06:22

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

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30 Jun 2016, 07:54

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.

Answer: D.

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.

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.

Answer: D.

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.
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