How many subsets of the set {1,2...15} have 2 odd numbers and any number of even numbers? I have no idea as to even how to approach this problem. I do know that i have to do something with the odd and even sets though.
any help would be appreciated!
edit: exactly 2 odd numbers
 A: Assuming the problem asks for exactly 2 odd numbers, you can choose $2$ odd numbers out of the $8$ that are present in the set, in $\binom{8}{2} = 28$ ways.  Then you can form any subset of the seven even numbers available; this can be done in $2^{7}=128$ ways, since each even number is eigther in or out of the subset.
The answer, then, is $$ 28\cdot 128$$
A: Hint: The subsets you are looking at are defined by which $2$ odd numbers they contain and by what subset of the even numbers they contain.  So the number of such sets is the number of ways you can choose $2$ odd numbers from the available $8$ odd numbers, times the number of subsets of the set of $7$ possible even numbers.
A: You have $8$ odd numbers in the set.  So you can have $\binom{8}{2}$ different combinations of odd pairs.
Now, you only have seven even numbers remaining to draw from.  You can have 0, 1, 2, ..., 7 even numbers in your subset, so you are looking at combinations of them.
So if I'm not mistaken, 
$$28\cdot \sum_{k=0}^7\binom{7}{k}=28\cdot2^7$$
