The usual name for such problems is "stars and bars": think of a list like this:
where each bar means "select this element" and each star means "omit this one", and you apply this rule to the sorted sequence of elements. What you do is look at "runs" of stars. Write down the length of each run, and when you're done, you have a sequence of nonnegative numbers that adds up to $k$.
See Stars and Bars Combinatorics for detailed answers on how to solve such problems.
As requested, a detailed example: Let's look at four items, in ascending order, and we pick two. The possible picks, represented with stars and bars, are
There are six of these. You can see, from this picture, that this also corresponds to a typical "put items in bins" problem by treating the bars as sides of the bins. In the first situation, two stars are in the left bin; in the next, there's a star in the left and middle bin, and so on.
The "second stars and bars theorem" tells us that for $n$ objects in $k$ bins, some of which may be empty, the number of choices is $n+k-1 \choose k-1$. In our case, we have 2 objects in 3 bins, and get $4 \choose 2 = 6$.
Now see if you can apply this to your problem.