How would you work out these combinations? 
*

*If there are 16 different ice-cream flavours, how many combinations are there for a two scoop? 

*If there are still 16 different ice-cream flavours, how many combinations are there for a three scoop?
How would you work out the above combinations? I found it just sitting in my notes app and I don't recall I ever found an answer.
My thoughts at the moment are 16 * (no. of scoops) but I am still lost since of course the flavours can appear in any order. (As my SE profile will demonstrate, I'm not much of a mathematician!)
So what would an equation be for the following as applicable to both Problem 1 and Problem 2?
A. Working out the number of combinations including duplicate scoops (e.g. chocolate-chocolate-vanilla)
B. Working out the number of combinations where a flavour only appears once in each possible combination (e.g. chocolate-vanilla-strawberry and then vanilla-chocolate-strawberry).
Any help would be much appreciated!
 A: 1: This problem is equivalent to putting two identical balls into 16 boxes (with repetition allowed).
$\binom{2+16-1}{2}$ 
Where 2 is the number of balls, and $16-1$ is the number of separators needed to divided a line segment into 16 smaller segments,each segment represents a box.
Each way of putting the balls (each combination of flavors) will be corresponds to a arrangement of $2$ balls and $16-1$ seperators (where balls and seperators are indistinguishable among themselves).
2.$\binom{3+16-1}{3}$.
A. As shown above, where we assume repetition is allowed.
B. If repetition is not allow,


*

*$\binom{16}{2}$

*$\binom{16}{3}$
A: 
1) working out the number of combinations including duplicate scoops
  (e.g. chocolate-chocolate-vanilla)

Consider the case where there is only one scoop of ice cream. There are 16 flavors (choices), and thus 16 "combinations." The next case is 2 scoops. One way to think about this problem is to consider how many choices you have per scoop. There are 16 choices for the first scoop and 16 choices for the second scoop since duplicates are allowed. This works out to $16^2 = 256$. It should be more clear as to how you can expand this to more scoops.

2) working out the number of combinations where a flavour only appears
  once in each possible combination (e.g. chocolate-vanilla-strawberry
  and then vanilla-chocolate-strawberry)

Now try the case of 3 scoops. There are 16 choices for the first scoop, but 15 choices for the second scoop since duplicates are not allowed. For the third scoop there are 14 choices, which works out to $16*15*14=3,360$. Note that with this way of combining flavors, the order is important since chocolate-vanilla-strawberry and vanilla-chocolate-strawberry are both counted.
EDIT (for cases where duplicates are allowed):

1.If there are 16 different ice-cream flavours, how many combinations are there for a two scoop?

There's a decent explanation from another post. The formula is indeed
$$
\binom{n+k-1}{k}
$$
where $n$ is the number of flavors and $k$ is the number of scoops. This is called a combination. Note that using this formula gives a different answer than I originally provided, since combinations do not consider the order in which objects are counted
$$
\binom{16+2-1}{2}=\binom {17}2=\frac{17\cdot16}{1\cdot2}=136
$$


  
*If there are still 16 different ice-cream flavours, how many combinations are there for a three scoop?
  

Using the formula above it can be worked out similarly
$$
\binom{16+3-1}{3}=\binom {18}3=\frac{18\cdot17\cdot16}{1\cdot2\cdot3}=816
$$
