Question: Hey I would like to know the correct approach to this problem. I think I'm close but I'm confusing myself. 

How many sets of size five are there from the natural numbers 1-100 that contain exactly two odd numbers?

First we would select 2 odd numbers (50 choose 2)

Then sum it with (50 choose 3)? For the even #s?

Answer: 
Not sum it, multiply it. This is because for **every way** of choosing the odds, there are $\binom{50}{3}$ ways to choose the evens.
Rule of product, not rule of sum. We choose our even numbers independently of the odd numbers. So $\binom{50}{2} \binom{50}{3}$. Note the rule of sum is for disjoint cases (ie., partitioning the elements you are counting).

Consider a smaller example- one even number and one odd number from $\{1, ..., 6\}$. So we can have: $(1, 2), (1, 4), (1, 6)$. We then apply the same pairs for $3, 5$. This gives us $9$ pairs, which is: $\binom{3}{1} \binom{3}{1}$, not $\binom{3}{1} + \binom{3}{1}$.