Amount of pairs of elements This is from a former Norwegian mathematics exam. I have solved a) and b). I have a kind of explanation for c), but I believe it's wrong. d) is supposedly connected, but I can't see why.
A handball player has n numbered balls that she shoots against a goal. She notes which balls that hit the goal and which ones miss.
a) How many possible results are there, and how many possible results is it with exactly k balls that hit?
My answer: 
Possible results $2^n$
Results with exactly k balls that hit: $\binom{n}{k}$
b) A goalkeeper is trying to stop the shots that hit the goal. She notes how many shots that go outside, how many are saved and how many are goals. How many results are possible for the attempt?
My answer: 
There are three possibilities for each n: outside, catch and goal. Because of that the amount of possible results is: $3^n$
c)Let S be a set with n elements and let R be the amount of subsets of S. Show that the amount of pairs (A,B) of elements in R, so that $A\subseteq B$ as subsets of S, is $3^n$
My answer:
There are three possibilities for each n:
1) n is not a subset of B
2) n is a subset of B but not A
3) n is a subset of both B and A
Because of this the amount of pairs is $3^n$
d) Show that:
$\sum_{j=0}^{n}2^j \binom{n}{j}=3^n$
My thoughts:
a) shows that the possible number of results is $2^n$ and that the number of results with exactly k balls that hit is $\binom{n}{k}$
Because of that 
$\sum_{k=0}^{n} \binom{n}{k}=2^n$
But I can't get from there to 
$\sum_{j=0}^{n}2^j \binom{n}{j}=3^n$
 A: In your discussion of part (c), I would use some symbol other than $n$ to denote a particular element of $S$.  You are correct that for each element $m \in S$ there are three mutually exclusive possibilities.  Either $m \notin A, m \in A \setminus B, \text{ or } m \in B$.  Assign one of these three states to each of the $n$ elements of $S$ and you get a unique ordered pair $(A. B)$ with $B \subseteq A$.  There are therefore $3^n$ such ordered pairs.
Use part (c) to solve part (d).  There are $\binom{n}{j}$ ways to choose subsets $A$ of size $j$ from $S$.  For each of these subsets $A$, there are exactly $2^j$ ways to choose a subset $B \subseteq A$.  Thus, the number of ordered pairs $(A, B)$ with $B \subseteq A$ is
$$\sum_{j=0}^n 2^j \binom{n}{j}=3^n,$$
where the last equality is precisely what you proved in part (c).
A: In question (c), you have a conflict of notation, when you say $n$ is not a subset of $B$. And then , you state that the number of pairs is $3^n$. You should edit this part by saying "for each number $k$", instead of "$n$". 
Moreover, an element is not a subset but of another set, it belongs to it (the mathematical notation is $\in$).
By editing this, I think that the overall argument is good.
For d) you should remark that to count the number of such pairs $(A,B)$ you can follow the following process :
For each $k \leq n$, you can first choose a subset $B$ of size $n$ (You have $\binom{n}{k}$ possibilities). And then for each subset $B$, you can choose any subset $A$ in $B$, hence you have ($2^k$ possible subset $A$ of $B$).
Can you conclude with that ?
