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Question 32 : In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers ?

- 16
- 20
- 14
- 15

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Given that 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers.

We have to find by how many number of ways the erasers are distributed such that each kid gets at least one eraser but no body gets more than 3 erasers.

Let us assume the 4 kids to be a , b , c , d such that a + b + c + d = 7.

It can be distributed in ^{6}C_{3} ways

⟹ \\frac{6 × 5 × 4}{3 × 2 × 1}) = 20

It is given that no body gets more than 3 erasers so let us say that one kid could get 4 , 5 , 6 eraser.

If one kid a, gets 4 erasers the other kids b , c , d can get each one respectively such that 4 , 1 , 1 , 1 or 1 , 1 , 1 , 4 or 1 , 4 , 1 , 1
or 1 , 1 , 4 , 1 so there are chances of 4 ways.

So, 20 – 4 = 16 ways

The question is **"In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers ?" **

Choice A is the correct answer

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