Discrete Math Pascal's Triangle and n choose k A juggling club has 5 male members and 4 female members.
a) I know the answer to this: "How many nonempty sets of juggling club members can be chosen to perform in a show" is 
Answer: 2^9 - 1.  However,
b) How many nonempty subsets of juggling club members containing at least one female can be chosen? Answer: Shouldn't it be 2^9 - 2^5 - 1?  My professor had the answer as just 2^9 - 2^5 without the minus one.  But we still need to exclude the empty set right?
Also how can we solve this using n choose k?  I am really starting to understand n choose k and it's relation with Pascal's Triangle, but I would like to apply it to this question here.
 A: $$\underbrace{(2^9 - 1)}_{\text{non-empty subsets}} - \underbrace{(2^5 - 1)}_{\text{non-empty sets of men only}} = 2^9 - 2^5$$
A: It should be the number of nonempty sets you can build with 9 elements, $\sum{9 \choose k}$ which is $2^9 - 1$ MINUS the number of nonempty sets you can build with 5 elements which would correspond to the number of sets in which there are no female jugglers, which would be $\sum{5 \choose k}$, which equals $2^5-1$
Your teacher is correct because the two "$-1$" compensate each other
If you didn't know about the $\sum {n \choose k} = 2^n$ I suggest you try to prove it by induction or you visit http://en.wikipedia.org/wiki/Binomial_coefficient
A: Although formulas are very useful, it is helpful to do a full analysis each time. 
There is a total of $2^9-1$ ways to choose a non-empty subset of our group. Pity about not counting the empty set: I would prefer the performance.
Call a choice of jugglers bad if it is all male. The same analysis shows there are $2^5-1$ bad choices.
Thus the number of good choices is $(2^9-1)-(2^5-1)$. This is $2^9-2^5$.
Or else we could have not worried about the empty set. There are $2^9$ ways to choose a subset, possibly empty. There are $2^5$ ways to choose a subset of the boys. So the number of subsets that contain at least $1$ girl (and are therefore non-empty) is $2^9-2^5$.
The binomial coefficients approach is unpleasant, so we compromise and use a mixed approach. 
$1$ girl: There are $\binom{4}{1}$ ways to choose her, and for each choice there are $2^5$ ways to choose a subset of the boys to keep her company. That gives a total of $\binom{4}{1}2^5$ choices. 
$2$ girls: There are $\binom{4}{2}$ ways to choose them, and $2^5$ ways to choose a subset of the boys to keep them company.
$3$ girls: There are $\binom{4}{3}$ ways to choose them, and  $\dots$.
$4$ girls: There are $\binom{4}{4}$ ways to choose them, and  $\dots$.
Add up.
