How many sets of size five are there from the natural numbers 1-100 that contain exactly two odd numbers? 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?
 A: 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.
A: 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 $\binom{3}{1} + \binom{3}{1}$. 
A: Choose 2 odd numbers from 50 : $\binom{50}{2}$
For each of these choices, you can choose 3 even numbers from the 50 even numbers : $\binom{50}{3}$  
Thus it should be $\binom{50}{2}.\binom{50}{3}$  
A: We want a set of size $5$, with $2$ odd numbers and $3$ even numbers. Converting this into math lingo, we get $\binom{50}{2} \cdot \binom{50}{3}$, and not their sum because we want both and not one or the other.
