I have $5$ flavors of ice cream and $6$ scoops available I've seen many questions similar to this but I have never seen one with more scoops available than flavors. 
Question:

I have $5$ flavors of ice cream and $6$ scoops available, how many different combinations can be produced. You are able to repeat as many flavors and order of the scoops does not matter.

 A: Here is another way to look at this: Imagine you had 4 bars and 6 dots. These 4 bars determine 5 regions. The number of dots in each region is the number of scoops of that flavor.
Ex: If you want two of the first flavor, two of the third flavor and two of the 5th flavor; you have ..||..||..
Then, your total number of combinations is just the number of ways to arrange 6 dots and 4 bars = 10 choose 4.
A: This is asking for the number of solutions to:
$$\sum_{i=1}^5 x_i = 6$$
where $0\le x_i\le6$.
In this $x_1$ is the number of scoops flavor $1$ and so on, so we consider $0+1+1+4+0$ as different to $1+0+4+0+1$ and so on.
The partitions of $6$ into at most $5$ parts are $6,51,42,411,33,321,3111,222,2211,21111$.
By adding with zeroes if necessary, we can count the number of compositions of each as
$$\dfrac{5!}{\prod_\limits{k=0}^6 (||=k)!}$$
where $(||=k)$ is the number of elements that equal $k$.
So the partition $6$ gives $\frac{120}{24}=5$ (from the $0$-count), corresponding to $60000,06000,00600,00060,00006$.
Continuing we get $20,20,30,10,60,20,10,30,5$, summing to $210$ with the previous $5$.
Which happens to be $\dbinom{10}{4}$.
