Why the cardinality of the two sets are equal?

Question

Problem. Consider the set $X=\{0,1,2,3\}$

1. List all subsets of $X$ which are disjoint from $\{0,1\}$.

2. List all subsets of $X$ which have $\{0,1\}$ as a subset.

3. How many subsets did you find in the previous two parts? What do you notice? Can you explain why this might be?

I have found that there are four subsets in part 1 and part 2. I am not sure how to explain it though (part 3)?

So in part 1, I have $P(X)\backslash\{\{0\},\{1\},\{0,1\},\{0,2\},\{0,3\},\{1,2\},\{1,3\},\ldots\}$ where $P(X)$ is the power set of $X$ and in part 2, I am finding all subsets $A$ of $X$ such that $A\cap\{0,1\}=\{0,1\}$ so $\{0,1\}\subseteq A$ but I yet can not see the reason clearly.

Answer 1

A subset in part 1 can be constructed as follows:

Take any subset $X'$ of $\{2,3\}$, set $X = X'.$

A subset in part 2 can be constructed as follows:

Take any subset $X'$ of $\{2,3\}$, set $X = X' \cup \{0,1\}$.