Xjyuk

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.

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: $5+12-1choose12$ = $16choose12$.

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.

1. 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 (°°|°°°°°°°°°°).


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


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



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




5. Now we have 2 balls left, which we put them in the 5th container.

In how many ways can we place the four balls in the 16 spaces we have?

The correct formula is $$binomn+k-1k-1=binomn+k-1n.$$

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 $$binom 5+22 = 21 text ways. : square$$

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 $$binomn+k-1k-1=binomn+k-1n$$ ways to do this.

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 $binom121$ or $binom120$?