Theorem: Let $A_1, A_2, \ldots, A_n$ be $n$ sets. Then, $ \left| \bigcup_{k = 1}^n A_k\right| = \sum_{S \subseteq [n] , S \neq \emptyset}^{ } (-1)^{|S|+1} \left|\bigcap_{i \in S} A_i\right| .$ Proof: We want to count every element in $\bigcup_{k = 1}^{n} A_k$ exactly once and count every other element $0$ times. Consider an arbitrary element $x \in \bigcup_{k=1}^{n} A_k$. Consider $\sum_{k = 1}^{n} - \left(\sum_{i,j \in [n], i < j} |A_i \cap A_j|\right)$. Let $K :=$ the number of elements containing $x$. We can see that the number of times $x$ is counted in $\sum_{i,j \in [n] \\ i < j} |A_i \cap A_j|$ is $\binom{K}{2}$ since only the two-wise intersections containing $x$ one from picking $2$ of the sets $K$ is in. Where does this come from? Consider an example when $x \in A_2, A_3, A_{10}$. Then, $K = 3$ and so there are $\binom{3}{2}$ ways to sets from among $A_2, A_3, A_{10}$ such that $x$ is in their intersection. We similarly find that the number of times $x$ is counted in the three-wise sets is $\binom{K}{3}$. We find that $x$ is counted $ \binom{k}{1}-\binom{k}{2}+\ldots \pm \binom{k}{n} = \binom{k}{1}-\binom{k}{2}+\ldots \pm \binom{k}{k} .$ From here, notice that $0= (-1+1)^{n} = \binom{k}{0} - \binom{k}{1}+\ldots \pm \binom{k}{k} = 1 - (\text{\# times $x$ is counted})$so the number of times $x$ is counted is $1$ so we are done. It turns out, the principle of Inclusion Exclusion can be used to solve the area of circular triangles...