In how many ways can you distribute $3$ chocolates among $2$ kids if you have to give all $3$ of the chocolates to the kids? In how many ways can you distribute $3$ chocolates among $2$ kids?
One kid can get none. But we need to give away all $3$ of the chocolates to the $2$ kids.
Why is it wrong to use $3C2$(which gives just one way) when we are distributing $3$ chocolates to $2$ kids?
Can you explain in simple words? Thank you.
 A: Let $x_1$ denote the number of chocolates that are given to the first child; let $x_2$ denote the number of chocolates that are given to the second child.  Since we must distribute all three chocolates to the two children, 
$$x_1 + x_2 = 3$$
The number of ways we can distribute the chocolates to the two children is equal to the number of solutions of the equation $x_1 + x_2 = 3$ in the nonnegative integers.  There are four such solutions.  They are $(0, 3)$, $(1, 2)$, $(2, 1)$, and $(3, 0)$.  
A: Here's another way to do it:
Denote the three chocolates with the symbols %%%, and consider an additional symbol |. 
Now move around those symbols, and interpret that the % which end up to the left of | are chocolates given to the first child, and the % which end up to the right of | are given to the second child. For example, one possible permutation is %%|%, which means the first child got 2 chocolates and the second one got 1. |%%% means the second child got all chocolates. 
There are $4!$ possible arrangements of the symbols %%%|, but since the $3$ chocolates are identical you need to divide by $3!$ to get all different orderings, which gives:
$${4! \over 3!}=4$$
This is also equivalent to ${4 \choose 1}$ since we're choosing a place for the | in a row formed by three % and a single |.
