Question

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?

Answer 1

The question is ver simple.

In order to solve this kind of questions one has to have good grasp over Permutation & Combination.

Well lets solve this.

There are 4 children and 7 erasers in total.

Each of them has to have atleast 1 eraser and no one can have more than 3 erasers.

Therefore, we can distribnute 4 erasers at first one to each of the children.

Now the remaining 3 erasers can be distributed among 4 children in:

4 * ( 4C3) ways = 4 * 4 = 16 ways.

Answer 2

This is a question on permutations and combinations.

Given that each kid must get 1 eraser we first give one to each.

The remaining erasers are :

7 - 4 = 3

For the three remaining we can distribute them in two ways :

The first is :

2,1, 0,0

This can be done in :

4!/2! = 12 ways

The second is :

1,1, 1,0

This can be done in :

4!/3! = 4 ways

The total ways is :

12 + 4 = 16 ways