TITLE: How many 19-bit strings can be generated from having an even number of ones?
QUESTION [8 upvotes]: So essentially how many 19-bit strings can you make with 2 1's or 4 1's or .... or 18 1's? 
I know the # of 19-bit strings that can be produced with 2 1's would be 19!/17!2! and the number of 19-bit strings that can be produced with 4 1's would be 19!/15!4! ..... up until 19!/18! in the case where there are 18 1's. The thing I don't understand is how much overlap occurs. 
I know this problem has to do with inclusion-exclusion principle, I am just confused on how to calculate the intersection of every single possible outcome.

REPLY [3 votes]: From the approach you took to the problem, my first thought was to use the fact that $\binom{19}{k}=\binom{19}{19-k}$; because $19$ is odd it follows that
$$\sum_{\substack{k=0\\k\text{ even}}}^{19}\binom{19}{k}
=\sum_{\substack{k=0\\k\text{ even}}}^{19}\binom{19}{19-k}
=\sum_{\substack{k=0\\k\text{ odd}}}^{19}\binom{19}{k}.$$
Adding the left hand side and right hand side together we get the sum over all $k$, so
$$2\times\sum_{\substack{k=0\\k\text{ even}}}^{19}\binom{19}{k}=\sum_{\substack{k=0\\k\text{ even}}}^{19}\binom{19}{k}+\sum_{\substack{k=0\\k\text{ odd}}}^{19}\binom{19}{k}=\sum_{k=0}^{19}\binom{19}{k},$$
and this equals $2^{19}$ by the binomial theorem, see also CR Drost's answer. So your answer is $2^{18}$.
But I like Doug M's answer much better.