Power Set Intersection Question 
Given the sets $X = \{1, 2, \dotsc, 10\}$ and $Y = \{1, 2, \dotsc, 12\}$,
  compute $|P(Y) \setminus P(X)|$, where $P$ is the Power Set
  operation. Explain your answer.

I would argue that the answer is $2^{12} - 2^{10}$.
All power sets of $X$ will be covered by $Y$ up to $10$.
 A: Your answer and idea is right, but $X$ will be covered by $Y$ up to $10$ is rather unprecise. I would formulate this by saying that because $X \subseteq Y$ we also have $P(X) \subseteq P(Y)$ and therefore
$$
 |P(Y) \setminus P(X)|
 = |P(Y)| - |P(X)|
 = 2^{12} - 2^{12}
 = 3 \cdot 2^{10}.
$$
Another way to see this is to notice that the elements of $P(Y) \setminus P(X)$ are precisely the subets of $Y$ which are not contained in $X$, i.e. the subsets of $Y$ containing $11$ or $12$.  To get all these subsets of $Y$ we take every subset $A$ of $X$, which there $|P(X)| = 2^{10}$ many of, and then take the unions $A \cup \{11\}$, $A \cup \{12\}$ and $A \cup \{11,12\}$. This results in the desired subsets of $Y$, which there then $3 \cdot 2^{10}$ many of.
A: So $S \in P(Y)$ is in $P(Y) \setminus P(X)$ iff either $11$ or $12$ are elements of $S$, else $S \in P(X)$. So $S \in P(Y) \setminus P(X)$ iff it follows one of the following three mutually exclusive conditions:
\begin{cases}
11 \in S & 12 \not \in S, \\
11 \not \in S & 12 \in S, \\
11 \in S & 12 \in S . 
\end{cases}
Thus we can write
$$ P(Y) \setminus P(X) = \{ \{11\} \cup F : F \in P(X) \} \cup \{ \{12\} \cup F : F \in P(X) \} \cup \{ \{11, 12\} \cup F : F \in P(X) \} , $$
where the union is disjoint. But we can also note that there is a clear bijection between $P(X)$ and each of the aforementioned sets, so $P(Y) \setminus P(X)$ can be written as the disjoint union of three sets of size $| P(X) |$, so \begin{align*}
|P(Y) \setminus P(X)| & = 3 |P(X)| \\
& = 3(2^{10}) \\
& = (4 - 1)2^{10} \\
& = 4(2^{10}) - 2^{10} \\
& = 2^{12} - 2^{10} .
\end{align*}
So you are correct, though you'd do well to better explain what your're getting at.
