How to count the amount of subsets within a set Let S be a set of size 37, and let x, y, and z be three distinct elements of S. How many subsets of S are there that contain x and y, but do not contain z?
(a) $2^{33}$
(b) $2^{34}$
(c) $2^{35}$
(d) $2^{37} − 2^{35} − 2^{36}$
This is a review question, I have no idea how to approach it, If someone can help that'd be appreciated!
My attempt is that $2^n$ determines the number of subsets in a set
So without Z, there is no {Z}, {Z,X}, {Z,Y}
therefore you take 3 off of 37 so $2^{34}$ is final answer
 A: We want to count the number of subsets of $S$, which contain $x,y$ but do not contain $z$. There is an obvious bijection between such subsets, and arbitrary subsets of $S\setminus\{x,y,z\}$ given by just adding $x$ and $y$ to a subset of $S\setminus\{x,y,z\}$. So the cardinality of $S\setminus\{x,y,z\} = 37-3 = 34$ and so there are $2^{34}$ subsets.
Another way of thinking about the bijection, is to say, to define a subset of $S$ we need to decide whether each element is in or out. We are told $x$ and $y$ are in, and also that $z$ is out. That leaves us $34$ more choices of "in" and "out" which is $2^{34} $ choices.
A: Fix $x$ and $y$ in the set and start adding elements one by one from from the rest of the 34 elements(without including $z$). 
There is one combination with just $x$ and $y$.
There are $34\choose 1$ combinations with 1 extra element, $34\choose 2$  combinations with two extra elements and so on.
So the total number of combinations is 
$$1+{34\choose 1} + {34\choose 2}+....{34\choose 34}=2^{34}$$ 
