Problem: Find the number of subsets of $\{1,2,3,4,5,6,7,8\}$ that are subsets of neither $\{1,2,3,4,5\}$ nor $\{4,5,6,7,8\}$. Solution: We want to find the cardinality of $\{A: A \not \subseteq \{ 1, \dots, 5, \} \text{ and } A \not \subseteq \{ 4,\dots, 8 \}\}$. But to count this, we can instead just deploy complementary counting, as $\begin{aligned} |\{A : A \not\subseteq \{1,\dots,5\} \text{ and } A \not\subseteq \{4,\dots,8\}\}| &= |\mathcal{P}(\{1,\dots,8\})| \\ &\quad - \left( |\{A : A \subseteq \{1,\dots,5\}\}| + |\{A : A \subseteq \{4,\dots,8\}\}| \right) \\ &\quad + |\mathcal{P}(\{4,5\})|. \end{aligned} $This sum is highly logical: * We first count the number of subsets from the grand set $\{ 1,\dots, 8 \}$ * We subtract all the "bad" cases (when $A \subseteq \{ 1, \dots, 5 \}$ or when $A \subseteq \{ 4, \dots, 8 \}$). * There is an overlap in this cases (namely when $A \subseteq \{ 4,5 \}$) so we add back thos cases we subtracted away twice. Computing each part is a small combinatorial argument in itself, and we get $2^8 - 2\cdot 2^5 + 2^2 = 196$.