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 $\dbinom n m$:

$\dbinom n m = \dfrac {n!} {m! \paren {n - m} }$

where $\dbinom n m$ is a binomial coefficient.

Thus the total number of subsets of $S$ whose cardinality is even is given by:

$\ds E = \dbinom n 0 + \dbinom n 2 + \dbinom n 4 + \cdots = \sum_{j \mathop \in \Z} \dbinom n {2 j}$

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

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

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

$\ds \sum_{j \mathop \ge 0} \dbinom n {2 j} = 2^{n - 1}$

$\blacksquare$