Let $E=\{1,2, \ldots, n\}$. Count the number of unordered pairs $(A, B)$ such that $A \cup B \subset E, A \cap B=\emptyset$ and $A \neq \emptyset \neq B$.
First, we count the number of ordered pairs ( $A, B$ ), included in $E$, allowing them to be empty. For this, to each function $f: E \rightarrow\{a, b, 0\}$ we associate the ordered pair $(A, B)$ such that $x \in A \Leftrightarrow f(x)=a$ and $x \in B \Leftrightarrow f(x)=b$. Since there are $3^{n}$ such functions, we have $3^{n}$ ordered pairs. To go from ordered pairs to unordered pairs, we subtract the pair $(\emptyset, \emptyset)$ and divide by $2: \frac{3^{n}-1}{2}$. To find the result, we subtract the pairs of type $(\emptyset, C)$, which are in number of $2^{n}-1$. Thus, the answer is $\frac{3^{n}+1}{2}-2^{n}$.