Mfcttrf

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.

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.

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?

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 $binom120$?