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

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17 Nov 2013, 07:10

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, 08:13

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, 23:04

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, 12:23

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

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21 Jun 2016, 07:22

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

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30 Jun 2016, 08:54

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.

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

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30 Jun 2016, 09:06

1

This post received KUDOS

Expert's post

Mariwa wrote:

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.

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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