How many ordered pairs $(A, B)$ are there, where $A$ and $B$ are subsets of a fixed $n$-element set and $A \subseteq B$?
I. solution. If $B$ has $k$ elements, then since a set with $k$ elements has a total of $2^{k}$ subsets, there are $2^{k}$ possibilities for set $A$. The $n$-element base set has $\binom{n}{k}$ subsets with $k$ elements, so the total number of $(A, B)$ pairs that meet the conditions of the problem is

$$
\sum_{k=0}^{n}\binom{n}{k} 2^{k}
$$

According to the binomial theorem, the value of this sum is exactly $(1+2)^{n}=3^{n}$. Therefore, the number of ordered $(A, B)$ pairs that satisfy the condition $A \subseteq B$ is $3^{n}$.

II. solution. Let $x$ be an arbitrary element of the $n$-element set in the problem. Since $A \subseteq B$, exactly one of the following three possibilities must hold, and the sets $A$ and $B$ uniquely determine which one:

- $x \in A$ and $x \in B$
- $x \notin A$ and $x \in B$;
- $x \notin A$ and $x \notin B$.

Indeed, if $x \in A$ and $x \notin B$, then $A$ is not a subset of $B$, so this is not possible. Conversely, if for all $n$ elements one of the above three possibilities holds, i.e., there is no element $x$ for which $x \in A$ but $x \notin B$, then $A \subseteq B$ also holds. The different elements can independently (in all possible ways) decide which of the three possibilities holds, and this already determines $A$ and $B$, so the answer to the problem is $3^{n}$.