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805+ (Hard)|   Combinations|      
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Five balls of different colors are to be placed in three different box 15 Nov 2013, 23:52
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adityapagadala
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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The n things need to be identical to use this formula.
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Five balls of different colors are to be placed in three different box 02 Feb 2014, 00:46
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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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Five balls of different colors are to be placed in three different box 30 Jun 2016, 08:54
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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.
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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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Five balls of different colors are to be placed in three different box 30 Jun 2016, 09:06
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Bunuel
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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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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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Five balls of different colors are to be placed in three different box 23 Jun 2020, 14:05
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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


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Solution:

First, let’s determine the number of ways 3 positive integers can add up to 5:

1 + 1 + 3 = 5 and 1 + 2 + 2 = 5

Notice the number of addends is the number of boxes and the sum is the number of balls.

We see that there are two options. Since the boxes are different, for each option above, there are 3!/2! = 3 ways to arrange the addends.

Furthermore, for the first option, we have 5 choices to put a ball in the first box and 4 choices to put another ball in the second box (then the last 3 balls must go to the last box). Therefore, for the first option, we have 5 x 4 = 20 ways to place the 5 balls in the 3 boxes (if we ignore the ways in which we can arrange the boxes (or addends) for the moment).

For the second option, we have 5 choices to put a ball in the first box and 4C2 = 6 choices to put 2 balls in the second box (then the last 2 balls must go to the last box). Therefore, for the second option, we have 5 x 6 = 30 ways to place the 5 balls in the 3 boxes (again, let’s ignore the ways in which we can arrange the boxes (or addends) for the moment).

Now, if we also take the arrangement of the boxes into consideration, then the total number of ways we can place 5 different colored balls in 3 different boxes is:

20 x 3 + 30 x 3 = 60 + 90 = 150

Answer: D
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Five balls of different colors are to be placed in three different box 17 Jun 2024, 06:58
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­1 - 2 - 2
There are 3 ways to choose the boxing containing 1 ball
There are 5 ways to choose such ball
For the 2nd box, there are 2C4 ways to choose 2 balls
Then only 1 way to choose the remaining balls for the last box
==> Total number of ways: 3 * 5 * 2C4 = 3 * 5 * 6 = 90

3 - 1 - 1
There are 3 ways to choose the box containing 3 balls
There are 3C5 ways to choose such 3 balls
For the 2nd box, there are 2 way to choose 1 ball for the box
Then only 1 way to choose the remaining balls for the last box
==> Total number of ways: 3 * 3C5 * 2 = 3 * 10 * 2 = 60

Total: 90 + 60 = 150
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Five balls of different colors are to be placed in three different box 29 Oct 2025, 21:08
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