Problem: Find the number of ordered triples of sets $\left(T_1, T_2, T_3\right)$ such that 1. each of $T_1, T_2$, and $T_3$ is a subset of $\{1,2,3,4\}$, 2. $T_1 \subseteq T_2 \cup T_3$, 3. $T_2 \subseteq T_1 \cup T_3$, 4. $T_3 \subseteq T_1 \cup T_2$. Solution: We can immediately see that if we are 'constructing' sets that satisfy these conditions, then we cannot ever let $x \in \{ 1,2,3,4 \}$ be only in one of $T_{1},T_{2},T_{3}$ (because then it cannot be a subset of the union of the others). So, for each element $x \in \{ 1,2,3,4 \}$ we have $5$ options, we can either let $x$ be in $T_{1},T_{2}$, in $T_{2}, T_{3}$, in $T_{3}, T_{1}$, in $T_{1},T_{2},T_{3}$, or none. Hence, there are $5^4 = 625$ ordered triples of sets satisfying the conditions.