TITLE: Number of even and odd subsets
QUESTION [5 upvotes]: Suppose we have the following two identities: 

$\displaystyle \sum_{k=0}^{n} \binom{n}{k} = 2^n$
$\displaystyle \sum_{k=0}^{n} (-1)^{k} \binom{n}{k} = 0$

The first says that the number of subsets of an $n$-set is $2^n$. The second says that the number of subsets of even size equals the number of subsets of odd size (of an $n$-set). Thus there are $2^{n-1}$ subsets of even length and $2^{n-1}$ subsets of odd length?
To combinatorially prove the second identity, let $A$ be a $k$-subset of $[n]$. Then note whether $k$ is odd or even?

REPLY [3 votes]: From another point of view you have:
$${\left( {1 + 1} \right)^n} =2^n= \sum\limits_{k = 0}^n {n \choose k}{{1^{n - k}}{1^k}}=\sum\limits_{k = 0}^n {n \choose k} $$
and
$${\left( {1 - 1} \right)^n} = 0 = \sum\limits_{k = 0}^n {n \choose k}{{1^{n - k}}{{\left( { - 1} \right)}^k}} = \sum\limits_{k = 0}^n {\left( { - 1} \right)}^k{n \choose k}$$