TITLE: 12 donuts split to 5 children
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.