lheiannie07 wrote:In how many ways can 6 chocolates be distributed among 3 children? A child may get any number of chocolates from 1 to 6 and all the chocolates are identical.

A) 10
B) 15
C) 21
D) 28
E) 56

We can apply the SEPARATOR METHOD, which I describe here:
https://www.beatthegmat.com/combinations-t120668.html

To guarantee that each child receives at least 1 chocolate, first give 1 chocolate to each child.
At this point, 3 of the 6 chocolates must still be distributed.
Thus, we need 3 identical chocolates and 2 identical separators, as follows:
OOO||.
The number of ways to arrange 5 elements composed of 3 identical identical chocolates and 2 identical separators = 5!/(3!2!) = 10.

The correct answer is A.

Alternate approach:

Let the 3 children be A, B and C.
To guarantee that each child receives at least 1 chocolate, first give 1 chocolate to each child.
Now WRITE OUT the different ways the remaining 3 chocolates can be distributed among the 3 children :
A =1 chocolate, B = 1 chocolate, C = 1 chocolate
A = 2 chocolates, B = 1 chocolate
A = 2 chocolates, C = 1 chocolate
B = 2 chocolates, A = 1 chocolate
B = 2 chocolates, C = 1 chocolate
C = 2 chocolates, A = 1 chocolate
C = 2 chocolates, B = 1 chocolate
A = 3 chocolates
B = 3 chocolates
C = 3 chocolates
Total ways = 10.

The correct answer is A.