Binomial Theorem past exam question, what do I do? I have trouble understanding what I'm supposed to do in some of these math questions. 
Here's an exam question from an old exam:
Let $A$ be a set with $n$ elements. The number of subsets of $A$ with $k$ elements is ${n \choose k}$, or:
$$ C(n,k) = \frac{n!}{k!(n - k)!} $$
Show that there are as many subsets having an odd number of elements as there are subsets with an even number of elements.
HINT: use the Binomial Theorem in the form:
$$ (1 + x)^n = \sum_{k = 0}^n {n \choose k} x^k$$
Then set $x = -1$.
Do I just choose an arbitrary number for $n$ then set $x = -1$ and work it out?
I don't know why I have such a problem understanding these questions. Thanks for your help. 
 A: You start with the set $A$ which cardinality is $n$. The number of subsets of $A$ with odd cardinalities is
$$
  \sum_{k:\text{ odd}}{n \choose k}
$$
and with even cardinalities is 
$$
  \sum_{k:\text{ even}}{n \choose k}
$$
so their difference is 
$$
  \sum_{k:\text{ even}}{n \choose k} - \sum_{k:\text{ odd}}{n \choose k} =   \sum_{0\leq k\leq n}{n \choose k}(-1)^k = (1+(-1))^n = 0
$$
A: If you set $x=-1$ in your formula, then
$$
0=(1+(-1))^n=\sum_{k=0}^n\binom{n}{k}(-1)^k,
$$
or splitting up the sum in positive and negative parts
$$
0=\sum_{\substack{k=0 \\k\text{ even}}}^n \binom{n}{k}+\sum_{\substack{k=0 \\k\text{ odd}}}^n\binom{n}{k}(-1),
$$
or equivalently
$$
\sum_{\substack{k=0 \\k\text{ odd}}}^n\binom{n}{k}=\sum_{\substack{k=0 \\k\text{ even}}}^n\binom{n}{k}.
$$
Note that 
$$
\sum_{\substack{k=0 \\k\text{ even}}}^n\binom{n}{k}=\binom{n}{0}+\binom{n}{2}+\cdots 
$$
is exactly the number ways to take out a subset of even size out of a set with $n$ elements and similarly with $k$ odd.
A: Hint: Don't use the hint. 
We assume (necessarily) that $n\ge 1$ and let $a$ be an element of $A$.
There is an obvious bijection between a) the odd sets containing $a$ and the even sets not containing $a$ (drop $a$ from the set) and b) the odd sets not containing $a$ and the even sets containing $a$ (add $a$ to the set).
Together this gives a bijection of odd and even sets, especially the numbers are the same.
