Finding subsets of a given size If we let $B$ be a subset of $A$, where $|A| = n, |B| = k$. What is the number of all subsets of $A$ whose intersection with $B$ has $1$ element? My task is to find a proof for this. This is my first proof based course and any guidance would be appreciated. 
Firstly, I know that since $B$ is a subset of $A$, $B$ contains all of $A$. For example, if:
  $$n = \{a, b, c\}$$
  $$k = \{a, b\}$$
then we know that $\{a\}, \{b\}$ are elements of intersection. Furthermore, $\{a, c\}$ and $\{b, c\}$ can also be included or excluded. I am just unsure on how to begin writing such a proof. Any input on where I can go from here would be great. Thanks.
 A: What you’ve written in the question indicates some confusions that ought to be addressed sooner rather than later.

Firstly, I know  that since $B$ is a subset of $A$, $B$ contains all of $A$.

This is backwards: $B\subseteq A$ means that $A$ contains every element of $B$, not that $B$ contains every element of $A$.

For example, if: $$n=\{a,b,c\}\\k=\{a,b\}$$ then we know that $\{a\},\{b\}$ are elements of intersection.

There are several problems here. First, $n$ and $k$ were already defined to be $|A|$ and $|B|$, respectively; what you probably mean here is that $A=\{a,b,c\}$ and $B=\{a,b\}$, in which case $n=3$ and $k=2$. Next, you’ve not mentioned any intersection, so it doesn’t make much sense to say that something is an element of it. I suspect that you mean $\{a,b,c\}\cap\{a,b\}$, but $\{a\}$ and $\{b\}$ are not elements of this intersection: the elements of the intersection are $a$ and $b$. The sets $\{a\}$ and $\{b\}$ are two of the four subsets of the intersection, the other two being $\varnothing$ and $\{a,b\}$. But if I’ve guessed right that you intended these two sets to be $A$ and $B$, then you’re not interested in this intersection anyway: first, since $B\subseteq A$, it’s automatic that $A\cap B=B$, and secondly, what you’re supposed to be doing is counting the subsets $S$ of $A$ such that $|S\cap B|=1$.
To get started, let’s use this example of $A=\{a,b,c\}$ and $B=\{a,b\}$ and actually tabulate those subsets. First, there are just two possibilities for the set $S\cap B$: the one element of this set can be $a$, in which case $S\cap B=\{a\}$, or it can be $b$, in which case $S\cap B=\{b\}$. If we add nothing else to $S$, we get the two sets $\{a\}$ and $\{b\}$, which are indeed subsets of $A$ having exactly one element in common with $B$. We can’t have both $a$ and $b$ in $S$, since $S\cap B$ would then contain more than one element, but we can add $c$ to either of these sets: the sets $\{a,c\}$ and $\{b,c\}$ also meet the requirement that they have only one element in common with $S$. And it’s not hard to see that these are the only four subsets of $A$ with that property: the other four subsets of $A$ are $\varnothing$ and $\{c\}$, which have no element in common with $S$, and $\{a,b\}$ and $\{a,b,c\}$, which have two elements in common with $S$.
Suppose that $A=\{a,b,c,d,e\}$ and $B=\{a,b\}$, so that this time $n=5$ and $k=2$. Once again, when we build a set $S\subseteq A$ such that $S\cap B$ contains exactly one element, there are $k=2$ choices for that element: it can be $a$ or $b$. Anything else has to come from the $n-k=3$ elements of $A\setminus B=\{c,d,e\}$:
$$\begin{array}{}
\{a\}&\{a,c\}&\{a,d\}&\{a,e\}&\{a,c,d\}&\{a,c,e\}&\{a,d,e\}&\{a,c,d,e\}\\
\{b\}&\{b,c\}&\{b,d\}&\{b,e\}&\{b,c,d\}&\{b,c,e\}&\{b,d,e\}&\{b,c,d,e\}
\end{array}$$
That makes a total of $2\cdot8=16$ sets.
In general it’s pretty clear that there will be $k$ ways to choose the one element of $S\cap B$. The rest of $S$ must be a subset of $A\setminus B$, and any subset of $A\setminus B$ will do.


*

*In terms of $n$ and $k$, how many subsets does $A\setminus B$ have?  

*Once you’ve found that number, how should you combine it with $k$ to get the answer to the question?

A: Hint:  you have to choose one element of $B$, plus some subset of $A \setminus B$
