How many 19-bit strings can be generated from having an even number of ones? 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. 

I like the accepted solution but I just wanted to say something about doing the mathematics the "hard" way. You know that there are $\binom m k = \frac{m!}{k!(m-k)!}$ ways to choose a set of $k$ items from a set of $m$ items. Those items can be numbered bits chosen to be 1s or 0s. Of course you know the identity that:
$$\sum_{k=0}^m \binom m k = 2^m$$ because of course if you sum up all the ways you might choose $k$ bits to be 1, then you get all possible bitstrings. However there is a simple proof of this which hinges on the idea that these coefficients appear in the binomial expansion, $$\sum_{k=0}^m \binom m k ~x^k ~y^{m-k}= (x + y)^m.$$ Just plug in $x = y = 1$ to find the above formula as a special case.
Well now you want to consider only even bitstrings, and so we can still "do it the hard way" by asking for a function of $k$ that is $1$ if $k$ is even or $0$ if $k$ is odd, and a great example is $[1 + (-1)^k]/2.$ Plugging that in, the sum that you want is just $$\sum_{k=0}^m \binom mk \frac{1 + (-1)^k}2 = \frac12 \sum_{k=0}^m \binom mk + \frac12 \sum_{k=0}^m \binom mk (-1)^k.$$ The first sum is clearly $2^m/2 = 2^{m-1}$ and the second sum we can use the above formula to reason that it is actually $(-1 + 1)^m/2 = 0^m/2 = 0.$ 
So you can do this the hard way if you'd prefer and you'll definitely get the same result.
