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Five balls of different colors are to be placed in three

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

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New post 15 Nov 2013, 18:38
when can we use (n-1)C(r-1) ?
It is equal to number of ways of dividing n things tor people where each can receive atleast 1? isnt it the case here?
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New post 15 Nov 2013, 22:52
adityapagadala wrote:
when can we use (n-1)C(r-1) ?
It is equal to number of ways of dividing n things tor people where each can receive atleast 1? isnt it the case here?


The n things need to be identical to use this formula.
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New post 16 Nov 2013, 00:05
oh got it...thanks for pointing out the difference..!!
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Re: Five balls of different colors are to be placed in three [#permalink]

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New post 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

Total 60+90 = 150


thank you so much for answering me.
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New post 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;

(2) Similarly: 5C2 x 3C2 x 1C1 x 3! / 2!=90;

do (1)+(2)=60+90=150
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New post 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
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Re: Five balls of different colors are to be placed in three [#permalink]

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New post 01 Feb 2014, 23:46
Leveraging off of Bunuel's approach, but attempting to tackle from a slightly different angle.

3-1-1:

Ball Combos-
(Choose 3 from 5)*(Choose 1 from 2 remaining)*(Choose 1 from 1 remaining)
5C3 * 2C1 * 1C1 = 20

Grouping Arrangements-
(Choose 2 "1 groupings" from 3 slots with the 3 grouping automatically filling the vacant spot)
3C2 = 3

20*3=60

2-2-1:

Ball Combos-
(Choose 2 from 5)*(Choose 2 from 3 remaining)*(Choose 1 from 1 remaining)
5C2 * 3C2 *1C1 = 30

Grouping Arrangements-
(Choose 2 "2 groupings" from 3 slots, with the 1 grouping automatically filling the vacant spot)
3C2 = 3

30*3=90

60+90=150
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New post 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.

What am I doing wrong?
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New post 30 Jun 2016, 08:06
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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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Re: Five balls of different colors are to be placed in three   [#permalink] 30 Jun 2016, 08:06

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