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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 ?

  1. 16
  2. 20
  3. 14
  4. 15

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Explanatory Answer

Method of solving this CAT Question from Permutation and Combination

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 6C3 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 ?"

Hence, the answer is 16

Choice A is the correct answer

 

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