Number of subsets of a set having r elements

You are confusing some things here. The argument given is the total number of subsets for a set of $n$ elements. Combinations, on the other hand, count the number of subsets of a given size.

For example, consider the set $\{1,\ 2,\ 3\}$. There are $2^3=8$ total subsets (of all sizes), which are $$\left\{\emptyset,\ \{1\},\ \{2\},\ \{3\},\ \{1,\ 2\},\ \{1,\ 3\},\ \{2,\ 3\},\ \{1,\ 2,\ 3\}\right\}$$ The subsets are symmetric in inclusion/exclusion (and this is why you only have two boxes: the first box represents inclusion and the second exclusion), for example the set $\{2\}$ has a complement, namely $\{1,\ 3\}$ where the first set is obtained by keeping $2$ while the second set is obtained by throwing away $2$.

Combinations count the subsets of a particular size ($n$ choose $r$ counts the number of $r$-element subsets of an $n$-element set). In our running example consider how many subsets of size $2$ there are: $$\left\{\{1,\ 2\},\ \{1,\ 3\},\ \{2,\ 3\}\right\}$$ for a total of $\binom{3}{2}$. The symmetry noted from before is also reflected in the fact that the binomial coefficients are symmetric $$\binom{n}{r} = \binom{n}{n-r}$$ which represents the fact that for every $r$-element subset that you keep, there's corresponding $n-r$-element subset that you've thrown away.

Of course the two concepts are intimately related. The total number of subsets is the sum of the number of subsets of every size. $$2^n = \sum_{r=0}^n\binom{n}{r}$$ This is in essense what the familiar binomial theorem states.