In how many ways can 7 identical erasers be distributed among 4 kids i

Each kid must get at least 1 eraser and all the erasers are identical. So there is only one way on which we can distribute 4 erasers to 4 kids.

Now, we are left with 3 erasers that need to be distributed among 4 kids.

Since no student can get more than 3 erasers, we can not give the remaining 3 erasers to one student only.

Therefore, 2 cases are possible:

CASE 1

1 eraser each is given to 3 students:

No. of possible ways = \(4C3\) = 4

CASE 2

1 eraser is given to 1 student and 2 erasers are given to another student:

No. of possible ways = \(4C1\) * \(3C1\) = (4*3) = 12

Total number of ways = 12 + 4 = 16

So the correct option is D

Asked: 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?

E |E |E |E

4 identical erasers are distributed among 4 kids and remaining 3 erasers are to be distributed among 4 kids

Number of ways = 6C3 = 20 ways
But no kid should get more than 3 erasers.
Therefore, the cases when remaining 3 erasers are distributed to 1 kids are to be excluded = 4 ways

Number of ways in which7 identical erasers can be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers = 20 - 4 = 16 ways

IMO D

There are 2 important equations for these types of arrangements


If the number of non negative integer solutions for the equation \(x_1 + x_2 \space + \space ..+ \space x_n \space = \space n\), then the number of ways the distribution can be done is:


(i) \(^{n+r−1}C_{r−1}\). In this case, value of any variable can be zero.


(ii) \(^{n−1}C_{r−1}\). In this case, minimum value for any variable is 1.




Given that: n = 7 erasers, r = 4 children and and each child should have at least one eraser, but the maximum should not exceed 3.

Total number of ways, where each child gets at least 1 eraser = \(^{n-1}C_{r-1} = \space ^{7-1}C_{4-1} = \space ^6C_3 = \frac{6 \space * \space 5 \space * \space 4}{3 \space * \space 2 \space * \space 1} = 20\)


These 20 also include the cases, where 1 child gets all 4 erasers. The total such cases is 4.

Therefore the required number of cases = 20 - 4 = 16



Option D

Arun Kumar

CrackverbalGMAT How do you know that total such cases is 4 ?  KarishmaB MartyMurray Can you please enlighten ?

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I think total such cases when a child gets more than 3 is 4 because after you distribute 4 identical chocolates among 4 children , you are left with 3 chooclates. The only way a child can get more than 3 chocolates is when these remaining 3 chocolates go into one particular child . There are 4 such children . Hence , there can be 4 ways in which a child can get more than 3 chocolates.
Say , there are 4 children such as A, B , C and D.
After each got 1 chocolate , either A can get all the remaining three ( thus making his count 1 + 3 = 4 which is more than 3 ) or B can get all the remaning three or C or D. Hence 4 such cases are possible.
KarishmaB , kindly check my reasoning.

This is a correct statement.

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