12 donuts split to 5 children Suppose you buy 12 identical donuts and wish to give them to the 5 children you are babysitting. How many different ways
can you distribute the donuts if:
(a) there are no restrictions
My answer for this question is, 
A) Combination with Repetition: (n+r-1,r) = (12+5-1,5) = (16,5)
B) But some people are getting different answers: (n+r-1,r) = (12+5-1,4)  = (16,4)
using the logic:
00|00|0000|000|0
there need to be 4 dividers to divide into 5 groups. 
Please tell me which one is the right answer.
 A: Let $x_1, x_2,...,x_5$ be the kids, so we could model this problem as $x_1+x_2+x_3+x_4+x_5=12$ (because the donuts are identical).
This can be solved with a combination with repetition: ${5+12-1}\choose{12}$ = ${16}\choose{12}$.
Edit:
Another way of thinking this is suppose we have to place 12 balls in 5 containers. The balls are identical and the containers are not. Say those (°°°°°°°°°°°°) are the 12 balls. 


*

*First we take 2 balls and place it in a container. Now we have 12-2
balls left. We are going to represent this as (°°|°°°°°°°°°°).

*Again we take 4 balls and we place them in the second container. Now
we have °°|°°°°|°°°°°°.

*Again we take another 3 balls and we place them in the 3rd
container: °°|°°°°|°°°|°°°.

*Again we take another ball and place it in the 4th container:
°°|°°°°|°°°|°|°°.

*Now we have 2 balls left, which we put them in the 5th container.
In how many ways can we place the four | in the 16 spaces we have?
A: The correct formula is $$\binom{n+k-1}{k-1}=\binom{n+k-1}{n}.$$

To understand why this is, consider $n$ indistinguishable balls, and we need to put them into $k$ groups. Now, imagine for each group, we can separate them with dividers. 
From here, we see we need $k-1$ dividers. Let's take a quick example:
How many ways are there to place $5$ indistinguishable balls into $3$ boxes?
Solution. We divide the groups by bars - for example, one possible way is $$**||***$$ As you can see, there are three groups:


*

*Group $1$: $2$ balls.

*Group $2$: $0$ balls.

*Group $3$: $3$ balls.


Another one would be like $$*|**|**$$ Experimenting with this a bit more, we see that we're just arranging $5$ balls and $2$ dividers, which can be done in $$\binom {5+2}2 = 21 \text{ ways}. \: \square$$

Now, for the general case:
We have to place $n$ balls in $k$ groups, which needs $k-1$ dividers. Then, using the example above, we conclude that there are $$\binom{n+k-1}{k-1}=\binom{n+k-1}{n}$$ ways to do this.
A: Sounds like you are struggling with a formula.  Maybe it will help to change the numbers so you can easily tell what the result will be.  Assume there is $1$ kid rather than $5$.  Is the correct number of ways to distribute $12$ donuts to $1$ child $\binom{12}1$ or $\binom{12}0$?
