BTGmoderatorDC wrote:In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 0 to 6 and all the chocolates are identical.

A) 21
B) 28
C) 56
D) 112
E) 224

Let the children be A, B and C. So A can get 0, B can get 0 and C can get 6 chocolates. Of course, this is different from A gets 6, B 0 and C 0, or, A gets 0, B 6 and C 0.

In the calculations below, we will show how 3 nonnegative integers can sum to 6 and the number of ways the 3 numbers can be rearranged among A, B and C (for example, the first calculation below describes the distribution of the 6 chocolates mentioned above):

0 + 0 + 6 = 6 ---> 3!/2! = 3 ways

0 + 1 + 5 = 6 ---> 3! = 6 ways

0 + 2 + 4 = 6 ---> 3! = 6 ways

0 + 3 + 3 = 6 ---> 3!/2! = 3 ways

1 + 1 + 4 = 6 ---> 3!/2! = 3 ways

1 + 2 + 3 = 6 ---> 3! = 6 ways

2 + 2 + 2 = 6 ---> 3!/3! = 1 way

Therefore, there are a total of 3 x 3 + 6 x 3 + 1 = 9 + 18 + 1 = 28 ways 6 chocolates can be distributed to 3 children.

Answer: B