number of subsets of even and odd Let $A$ be a finite set. Prove or disprove: the number of subsets of $A$ whose size is even is equal to the number of subsets of $A$ whose size is odd. 
Example: $A = {1,2}$. The subsets of $A$ are {},{1},{2}, and {1,2}.  Since there are two subsets of odd size ({1} and {2}) and two subsets of even size ({} and {1,2}) the claim holds for this particular example.

The number of subsets of size $k$ in a set with $n$ elements is $\binom{n}{k}$.
The Binomial Theorem states that 
$$(a + b)^n = \sum_{k = 0}^{n} \binom{n}{k}a^{n - k}b^k$$
In particular, if we let $a = b = 1$, we obtain
$$2^n = (1 + 1)^n = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}1^k = \sum_{k = 0}^{n} \binom{n}{k}$$
Therefore, the number of subsets of a set with $n$ elements is $2^n$. Furthermore, if we set $a = 1$ and $b = -1$ in the Binomial Theorem, we obtain
\begin{align*}
0 & = 0^n\\
  & = [1 + (-1)]^n\\
  & = \sum_{k = 0}^{n} \binom{n}{k}1^{n - k}(-1)^k\\
  & = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \cdots + (-1)^{n}\binom{n}{n}\\
  & = \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} - \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1}
\end{align*} 
where $\lfloor x \rfloor$ represents the largest integer less than or equal to $x$.  Thus, 
$$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} = \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1}$$
Hence, the number of subsets of a set with $n$ elements that contain an even number of elements is equal to the number of subsets that contain an odd number of elements.  
Moreover, since a set with $n$ elements has $2^n$ subsets and the number of subsets with an even number of elements is equal to the number of subsets with an odd numbers of elements, the set has $\frac{1}{2} \cdot 2^n = 2^{n - 1}$ subsets with an even number of elements and $2^{n - 1}$ subsets with an odd number of elements.
