Multichoose confusion; stars and bars Problem 1: Five people have 20 ice cream cones. In how many ways can the ice cream cones be distributed among the five people?
Answer: 
$${20+5-1\choose 5-1}= {24\choose 4}=10 626.$$ 
Problem 2: In how many ways can you choose 4 ice cream cones out of 10 if you can take the same ice cream cone repeatedly and the order doesn't matter?
Answer: 
$${13\choose 9}= {13*12*11*10\choose 4!} ={13\choose 4}=715.$$ 
Somehow they're kind of the same problem. How? I don't understand.
I would think that for the second problem there are 10 options for the first ice cram cone, 10 for the second, 10 for the third, 10 for the fourth, resulting in 10^4 options. Since order doesn't matter we can divide that by 4! getting 10^4/4! but that's wrong.
https://www.statisticshowto.datasciencecentral.com/multiset/ says stars and bars can be used for multichoose problems. 
Multichoosing (Stars and bars) says :
$\binom {n+k-1}{k-1}$ = $\binom {n+k-1}{n}$
 A: In order to see how the problems are the "same", you have to abstract them.
One abstraction that works is to imagine you have $k$ objects of some kind, and you need to take an action of some kind $n$ times.
Each action acts on one of the objects, and you can act on the same object once, more than once, or not at all.
Moreover, in the end we don't care about the order in which the actions were taken, just how many times an action was taken on each object.
I took particular care here to match the notation from the last formula in your question; $k$ and $n$ in the previous paragraph are the same $k$ and $n$ that occur in the formula $\binom {n+k-1}{k-1}$.
In problem 1 the objects are persons, and the action is to give an ice cream cone to that person. There is an unspoken assumption here that nobody cares which ice cream cone(s) they have, so if persons 1 and 2 were to trade one cone for one cone it would not count as a different way for the cones to be distributed.  (I think it is actually a flaw in the problem statement that this is not explicitly stated;
merely inserting the word "indistinguishable" before "ice cream cones" would be sufficient.)
In problem 2 the objects are ice cream cones, and the action is to choose a cone.
Note that in this problem the cones are distinguishable; otherwise we would have no way to count how many times each cone was chosen.
A more purely mathematical abstraction is that we are making a list of $k$ non-negative integers whose sum must be $n.$
In problem 1 the $m$th integer in the list represents the number of cones received by the $m$th person.
In problem 2 the $m$th integer represents the number of times the $m$th cone was chosen.
The reason the answer to problem 2 is not $10^4/4!$ is that there are not always $4!$ different orders in which you can make a given set of choices.
Yes, if the choices are cone 1 once, cone 2 once, cone 5 once, and cone 7 once, then there are $4!$ orders in which to make those choices.
But if the choices are cone 1 all four times, there is only one possible order: cone 1, then cone 1, then cone 1, then cone 1;
and this happens only once in the $10^4$ ways of making ordered choices.
By dividing $10^4$ by $4!$, you are saying that "cone 1 four times" only counts as $1/24$ of a way to choose cones.
By the way, the result of your proposed method is $10^4/4! = 1250/3 = 416.666\ldots,$ which is not even an integer.
