Count of Subsets with Even Cardinality/Proof 1

Theorem
Let $S$ be a set whose cardinality is $n$.

Then the number of subsets of $S$ whose cardinality is even is $2^{n-1}$.

Proof
Let $E$ be the total number of subsets of $S$ whose cardinality is even.

From Cardinality of Set of Subsets, the number of subsets of $S$ with $m$ elements is $\displaystyle \binom n m$:
 * $\displaystyle \binom n m = \frac{n!}{m! \left({n-m}\right)}$

where $\displaystyle \binom n m$ is a binomial coefficient.

Thus the total number of subsets of $S$ whose cardinality is even is given by:
 * $\displaystyle E = \binom n 0 + \binom n 2 + \binom n 4 + \cdots = \sum_{j \mathop \in \Z} \binom n {2j}$

Note the loose limits of the summation sign: the expression truly ranges over all $\Z$.

This is because, when $2 j < 0$ and $2 j > n$, $\displaystyle \binom n {2j} = 0$ by definition of binomial coefficient.

The result then follows from Sum of Even Index Binomial Coefficients:


 * $\displaystyle \sum_{j \mathop \ge 0} \binom n {2 j} = 2^{n-1}$