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Five balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty?
A. 100
B. 125
C. 150
D. 175
E. 200
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A. 100
B. 125
C. 150
D. 175
E. 200
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Updated on: 22 Feb 2025, 03:59
Originally posted by GMATinsight on 23 Feb 2021, 04:57.
Last edited by Bunuel on 22 Feb 2025, 03:59, edited 2 times in total.
Last edited by Bunuel on 22 Feb 2025, 03:59, edited 2 times in total.
Bunuel
Case 1: {1, 1, 3} two boxes with one ball and one box with 3 balls
Select box which gets three balls = \(^3C_1 = 3\)
Select 3 balls for that box = \(^5C_3 = 10\)
Remaining two balls to be put into two boxes in 2! = 2 ways
Desired outcomes = 3*10*2 = 60 ways
Case 2: {1, 2, 2} two boxes with two balls each and one box with 1 balls
Select boxes which gets two balls each = \(^3C_2 = 3\)
Select 2 balls for first of those selected box = \(^5C_2 = 10\)
Select 2 balls for Second of those selected box = \(^3C_2 = 3\)
Remaining one balls to be put into last box = 1 ways
Desired outcomes = 3*10*3*1 = 90 ways
My Answer = 60+90 = 150
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General Discussion
sujoykrdatta
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Expert reply
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Posts: 598
Bunuel
Possible distributions are:
3-1-1:
Choose the balls: 5c3 x 2c1 x 1c1 = 20
Decide the arrangements: 3!/2! = 3
Total = 20*3 = 60
Or
2-2-1
Choose the balls: 5c2 x 3c2 x 1c1 = 30
Decide the arrangements: 3!/2! = 3
Total = 30*3 = 90
Total = 60+90 = 150
Alternatively:
Each ball has 3 options (boxes)
Total ways = 3^5 = 243
Cases to be removed:
1. Cases where all balls are in 1 box = 3
2. Cases where all balls are in 2 boxes:
Choose 2 boxes = 3c2 = 3
Each ball has 2 options = 2^5 = 32
But, of these, there are 2 cases when the balls are in only 1 box; these we should ignore as we considered this in 'Case 1'; ie. 32-2 = 30
So total = 3c2 x (32-2) = 90
Hence, required answer
= 243 - 3 - 90 = 150
Bunuel - Please check the options once
Posted from my mobile device
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ScottTargetTestPrep
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Expert reply
Posts: 22,740
Bunuel
Solution:
Since each box must have at least one ball, we see that there are only two cases to consider:
Case 1: (3, 1, 1) or (1, 3, 1) or (1, 1, 3)
For (3, 1, 1), we have 5C3 x 2C1 x 1C1 = 10 x 2 x 1 = 20 ways to choose the balls.
For (1, 3, 1), we have 5C1 x 4C3 x 1C1 = 5 x 4 x 1 = 20 ways to choose the balls.
For (1, 1, 3), we have 5C1 x 4C1 x 3C3 = 5 x 4 x 1 = 20 ways to choose the balls.
Case 1 yields a total of 20 + 20 + 20 = 60 ways.
Case 2: (1, 2, 2) or (2, 1, 2,) or (2, 2, 1)
For (1, 2, 2), we have 5C1 x 4C2 x 2C2 = 5 x 6 x 1 = 30 ways to choose the balls.
Note that the two remaining choices (2, 1, 2,) or (2, 2, 1) yield 30 ways and 30 ways, respectively.
Thus, Case 2 yields 30 + 30 + 30 = 90 ways.
The total number of ways to place the balls such that no box remains empty is 60 + 90 = 150.
Answer: C
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We are distributing 5 different balls into 3 different bins.
In this scenario, which ball is selected and the ordering of which bin it enters both matter.
Since each bin must receive at least 1 ball (“no bin can be empty”), there are only 2 scenarios in which the balls can be distributed:
[2 - 2 - 1]
Or
[1 - 1 - 3]
Scenario 1: first distribute the 5 different balls into stacks of 3 - 1 - 1
No. of ways this can be done is:
5! / (3! * 1! * 1!)
Because 2 of the stacks will have identical amounts of balls, each unique division will be over counted by 2.
5! / (3! * 1! * 1!) * (1 / 2!)
Now, we need to find the no. of ways we can arrange or shuffle around the 3 stacks such that they are placed in the 3 different bins.
Again, since all the balls are different and the bins are different, the no. of different arrangements is 3!
5! / (3! * 1! * 1!) * (1 / 2!) * (3!) =
60 ways
OR
Scenario 2: [1 - 2 - 2]
Following the same logic as above, remembering that we again have 2 stacks of identical amounts of 2 and 2:
No. of ways = 5! / (2! * 2! * 1!) * (1 / 2!) * (3!) =
90 ways
60 + 90 = 150 ways
Posted from my mobile device
In this scenario, which ball is selected and the ordering of which bin it enters both matter.
Since each bin must receive at least 1 ball (“no bin can be empty”), there are only 2 scenarios in which the balls can be distributed:
[2 - 2 - 1]
Or
[1 - 1 - 3]
Scenario 1: first distribute the 5 different balls into stacks of 3 - 1 - 1
No. of ways this can be done is:
5! / (3! * 1! * 1!)
Because 2 of the stacks will have identical amounts of balls, each unique division will be over counted by 2.
5! / (3! * 1! * 1!) * (1 / 2!)
Now, we need to find the no. of ways we can arrange or shuffle around the 3 stacks such that they are placed in the 3 different bins.
Again, since all the balls are different and the bins are different, the no. of different arrangements is 3!
5! / (3! * 1! * 1!) * (1 / 2!) * (3!) =
60 ways
OR
Scenario 2: [1 - 2 - 2]
Following the same logic as above, remembering that we again have 2 stacks of identical amounts of 2 and 2:
No. of ways = 5! / (2! * 2! * 1!) * (1 / 2!) * (3!) =
90 ways
60 + 90 = 150 ways
Posted from my mobile device
See where am i wrng?
For (3,1,1)
Choosing 3 out of 5 balls--> 5c3
Choosing 1 out of 2 balls --> 2c1
Choosing 1 out of 1 --> 1c1
Now for a particular combination (b1 b2 b3 , b4, b5)
Main point: For a single event suppose, Triplets have three options of sitting in any one box so it chooses any one Box , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways ?
So, 5c3 * 2c1 * 1c1 * 3! .
What is the mistake that am i not able to catch?
For (3,1,1)
Choosing 3 out of 5 balls--> 5c3
Choosing 1 out of 2 balls --> 2c1
Choosing 1 out of 1 --> 1c1
Now for a particular combination (b1 b2 b3 , b4, b5)
Main point: For a single event suppose, Triplets have three options of sitting in any one box so it chooses any one Box , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways ?
So, 5c3 * 2c1 * 1c1 * 3! .
What is the mistake that am i not able to catch?
Saumyojit
Saumyojit
For (3,1,1)
Choosing 3 out of 5 balls--> 5c3
Choosing 1 out of 2 balls --> 2c1
Choosing 1 out of 1 --> 1c1
Choosing 3 out of 5 balls--> 5c3
Choosing 1 out of 2 balls --> 2c1
Choosing 1 out of 1 --> 1c1
Here you need to also identify the box which is getting 3 balls because it can be in these three ways (311)(131)(113)
You can calculate it in \(^3C_1\) ways and not in 3! ways
so the calculation of this case will be = \(^5C_3* ^2C_1*^ 1C_1*\)\(^3C_1\) = 60
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Your reply shows that you did not understand what my doubt is .
For a single event suppose, Triplets have three options of sitting in any one box so it chooses any one Box , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways .
Now what you are saying is that Triplets have three options of sitting in any one box so it chooses any one Box --> 3 ways .
But what about the single ball (B4 , B5)
I identified the box which is getting 3 balls . What are u talking about.
Read the above part again
For a single event suppose, Triplets have three options of sitting in any one box so it chooses any one Box , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways .
Now what you are saying is that Triplets have three options of sitting in any one box so it chooses any one Box --> 3 ways .
But what about the single ball (B4 , B5)
I identified the box which is getting 3 balls . What are u talking about.
Read the above part again
box1 box2 box3
b1b2b3 , b4 , b5
b1b2b3 , b5 , b4
are these two different way or one way?
b1b2b3 , b4 , b5
b1b2b3 , b5 , b4
are these two different way or one way?
Concept that was used first : Division of Objects into equal and unequal groups
This is where I was mistaking.
5c3 --> B1 B2 B3 chosen out of 10 combo
2c1--> b4 chosen out of 2 combo (b4 or b5)
1c1---> B5
Here i was dividing 5 balls in 3 groups **not** boxes in 3:1:1
Now, group 2 and group 3 is the same thing only .
(B1B2B3) , B4 ,B5 and (B1B2B3) , B5 ,B4 --> These two are same . SO, 2c1 is overcounting .
so, final expression of Division into Groups = (5c3 * 2c1 * 1c1 )/ 2!
So, Now comes the distribution part -->
For a event suppose, Triplets have three options of sitting in any one box so it chooses one , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways .
3options * 2options * 1 option
triplets single another single
So, Final expression of (3,1,1) = (5c3 * 2c1 * 1c1 ) * 3
This is where I was mistaking.
5c3 --> B1 B2 B3 chosen out of 10 combo
2c1--> b4 chosen out of 2 combo (b4 or b5)
1c1---> B5
Here i was dividing 5 balls in 3 groups **not** boxes in 3:1:1
Now, group 2 and group 3 is the same thing only .
(B1B2B3) , B4 ,B5 and (B1B2B3) , B5 ,B4 --> These two are same . SO, 2c1 is overcounting .
so, final expression of Division into Groups = (5c3 * 2c1 * 1c1 )/ 2!
So, Now comes the distribution part -->
For a event suppose, Triplets have three options of sitting in any one box so it chooses one , then there are two boxes left for 2 Single Distinct balls ; so ball 4 have 2 options so suppose b4 sits in any ONE out of 2 box and the remaining one box can be taken by b5 . So , 6 ways .
3options * 2options * 1 option
triplets single another single
So, Final expression of (3,1,1) = (5c3 * 2c1 * 1c1 ) * 3
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The best way to solve Distince to Distince is using inclusion exlusion principle.
All distribuion incluing zero balls allowed - (All distribuion incluing zero balls allowed in 3 boxes) + (All distribuion incluing zero balls allowed in 1 box)
= 3^5 − 3C1 × 2^5 + 3C2 × 1^5 = 243 − 96 + 3 = 150
All distribuion incluing zero balls allowed - (All distribuion incluing zero balls allowed in 3 boxes) + (All distribuion incluing zero balls allowed in 1 box)
= 3^5 − 3C1 × 2^5 + 3C2 × 1^5 = 243 − 96 + 3 = 150
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