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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?
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?
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