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Permutation and Combination- Distribution of Balls into Boxes

Learn the simple trick to master the tricky concept of Permutation and Combination: Distribution of balls into boxes.

In this article, we are going to learn how to calculate the number of ways in which x balls can be distributed in n boxes. This is one confusing topic which is hardly understood by students. But once mastered, it is the easiest topic of Permutation and Combination.

There can be 4 cases pertaining to this problem.
Case 1: Balls are same; boxes are same
Case 2: Balls are same; boxes are different
Case 3: Balls are different; boxes are same
Case 4: Balls are different; boxes are different
To understand it better, let's take an example.

Example: What is the number of ways in which you can distribute 5 balls in 3 boxes when:
• Balls are same; boxes are same
• Balls are same; boxes are different
• Balls are different; boxes are same
• Balls are different; boxes are different

Solution:

• When Balls are same; boxes are same
•  Group Permutation of balls (Number of ways of grouping) Ways of Distribution of boxes Total number 0,0,5 1 1 1*1=1 0,1,4 1 1 1*1=1 0,2,3 1 1 1*1=1 1,1,3 1 1 1*1=1 1,2,2 1 1 1*1=1

Hence, total number of ways = 1+1+1+1+1=5.

• Balls are same; boxes are different
•  Group Permutation of balls(Number of ways of grouping) Ways of Distribution of boxes Total number 0,0,5 1 3!/2!=3 1*3=3 0,1,4 1 3!=6 1*6=6 0,2,3 1 3!=6 1*6=6 1,1,3 1 3!/2!=3 1*3=3 1,2,2 1 3!/2!=3 1*3=3

Hence, total number of ways = 3+6+6+3+3=21.

• Balls are different; boxes are same
•  Group Permutation of balls(Number of ways of grouping) Ways of Distribution of boxes Total number 0,0,5 1 1 1*1 0,1,4 5C1*4C4=5 1 5*1=5 0,2,3 5C2* 3C3=10 1 10*1=10 1,1,3 5C3* (2!/2)=10 1 10*1=10 1,2,2 5C1 *(4!/(2!*2!*2)=15 1 15*1=15

Hence, total number of ways = 1+5+10+10+15=41

• Balls are different; boxes are different
•  Group Permutation of balls(Number of ways of grouping) Ways of Distribution of boxes Total number 0,0,5 1 3!/2!=3 1*3=3 0,1,4 5C1 = 5 3!=6 5*6=30 0,2,3 5C2 = 10 3!=6 10*6=60 1,1,3 5C3* (2C1/2) = 10 3!=6 10*6=60 1,2,2 5C1 *(4!/2!*2!*2)=15 3!=6 15*6=90

Hence, total number of ways = 3+30+60+60+90= 243.

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