What is the average cardinality of a subset of $\{1,2, \ldots, n\}$?
First method (based on an idea from students): We group a set with its complement. If $A \subset \{1,2, \ldots, n\}$, we denote $A^{c}$ the set of integers in $\{1,2, \ldots, n\}$ that are not in $A$. Let $N$ be this average cardinality. Then:

\[
\begin{aligned}
N & =\frac{1}{2^{n}} \sum_{A \subset \{1,2, \ldots, n\}} \operatorname{Card}(A)=\frac{1}{2^{n}} \sum_{A \subset \{1,2, \ldots, n\}} \frac{\operatorname{Card} A+\operatorname{Card}\left(A^{c}\right)}{2} \\
& =\frac{1}{2^{n}} \sum_{A \subset \{1,2, \ldots, n\}} \frac{n}{2}=\frac{1}{2^{n}} \cdot \frac{n}{2} 2^{n}=\frac{n}{2} .
\end{aligned}
\]

Other methods: We need to evaluate the sum

\[
S_{n}=\frac{1}{2^{n}} \sum_{k=0}^{n} k\binom{n}{k}
\]

If you don't know how to start, study the first few cases: $n=1,2, \ldots$ You will always find $S_{n}=n / 2$. Try to prove this.

Second method: If you have no idea, you can proceed by induction on $n$.

Third method, more advanced. Consider the polynomial $P_{n}(x)=\sum_{k=0}^{n}\binom{n}{k} x^{k}$. According to the binomial theorem, $P_{n}(x)=(1+x)^{n}$. Differentiate this equality with respect to $x$:

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

Evaluate this quantity at $x=1$:

\[
\sum_{k=0}^{n} k\binom{n}{k}=n 2^{n-1}
\]

The result follows.