Question :- Five balls needs to be placed in three boxes. Each box can hold all the five balls. In how many ways can the balls be placed in the boxes if no box can be empty and all balls and boxes are different
Please Help to solve this type of division and distribution combination sums
I approached like below
Case A :- 1,1,3
From 5 distinct balls i can group them into (1,1,3) in 5!/3!*2! no of ways
After grouping each group can be put to 3! ways
So together it is (5!/3!*2!)*3!
= 60 ways
Case B :- 1,1 2
From 5 distinct balls i can group them into (1,1 2) in 5!/2!*2! no of ways
After grouping each group can be put to 3! ways
So together it is (5!/2!*2!)*3!
= 180 ways
Case A + Case B
= 60 + 180 = 240 ways
Is my approach correct?
Five Distinct balls needs to be placed in three boxes
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Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marbleFive balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we we place the balls so that no box remains empty?
A. 150 B. 10 C. 60 D. 300 E. 375
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.
Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.
Total ways = 60+90 = 150.
The correct answer is A.
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Thanks a Ton Thus solved My confusionGMATGuruNY wrote:Case 1: 1 box has 3 marbles, the other 2 boxes each have 1 marbleFive balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we we place the balls so that no box remains empty?
A. 150 B. 10 C. 60 D. 300 E. 375
Number of box options for the box with 3 marbles = 3. (Any of the 3 boxes.)
For this box, the number of ways to choose 3 marbles from 5 options = 5C3 = (5*4*3)/(3*2*1) = 10.
Number of marble options for the next box = 2. (Either of the 2 remaining marbles.)
Number of marble options for the last box = 1. (Only 1 marble left.)
To combine these options, we multiply:
3*10*2*1 = 60.
Case 2: 1 box has 1 marble, the other 2 boxes each have 2 marbles
Number of box options for the box with 1 marble = 3. (Any of the 3 boxes.)
Number of marbles that could be placed in this box = 5. (Any of the 5 marbles.)
From the 4 remaining marbles, the number of ways to choose 2 marbles for the next box = 4C2 = (4*3)/(2*1) = 6.
From the 2 remaining marbles, the number of ways to choose 2 marbles for the last box = 2C2 = (2*1)/(2*1) = 1.
To combine these options, we multiply:
3*5*6*1 = 90.
Total ways = 60+90 = 150.
The correct answer is A.
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This isn't really a GMAT topic (marbles in boxes), but if you're curious, there are a few easy formulas that address the basic cases.