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.
Best Answer
One way to see this combinatorially is to consider the $10$ elements in some order. With each of the first $9$ elements, you have a free choice of either including it or not including it, for $2$ choices each. However, there's no choice for the $10$th element, since to ensure the total # of elements in the set is odd, this element must be included if the total # so far is even, and not included if the total # is odd. Thus, since there are $2$ choices for each of the first $9$ elements, this means there are $2^9$ choices overall.