Let $M$ be a  set with $n$ elements. How many pairs $(A, B)$ of subsets of $M$ are there such that $A$ is a subset of $B?$
1. **Understanding the Problem:**
   We need to find the number of pairs \((A, B)\) of subsets of a set \(M\) with \(n\) elements such that \(A \subseteq B\).

2. **First Solution (s1):**
   - We can use the binomial theorem to solve this problem.
   - For each element in \(M\), there are three choices:
     1. The element is in neither \(A\) nor \(B\).
     2. The element is in \(B\) but not in \(A\).
     3. The element is in both \(A\) and \(B\).
   - Therefore, for each element, there are 3 choices, and since there are \(n\) elements, the total number of pairs \((A, B)\) is \(3^n\).

3. **Second Solution (s2):**
   - Consider the set \(M\) with \(n\) elements.
   - For any pair \((A, B)\) for \(n\), it can be mapped to three pairs:
     1. \((A, B)\)
     2. \((A, B \cup \{n+1\})\)
     3. \((A \cup \{n+1\}, B \cup \{n+1\})\)
   - This means the number of such pairs becomes three times as much when we add an element.
   - For \(n = 0\), the answer is 1 because \((\emptyset, \emptyset)\) is the only pair.
   - Therefore, the answer is \(1 \times 3^n = 3^n\).

4. **Verification using Binomial Theorem:**
   - The sum \[2^0 \cdot \binom{n}{0} + 2^1 \cdot \binom{n}{1} + 2^2 \cdot \binom{n}{2} + \cdots + 2^n \cdot \binom{n}{n}\] can be interpreted as follows:
     - Each term \(2^k \cdot \binom{n}{k}\) represents the number of ways to choose \(k\) elements to be in \(B\) and then choose any subset of these \(k\) elements to be in \(A\).
   - By the binomial theorem, this sum equals \((2+1)^n = 3^n\).

\[
\sum_{k=0}^{n} 2^k \binom{n}{k} = (2+1)^n = 3^n
\]

Conclusion:
\[
\boxed{3^n}
\]