Problem: Define an ordered triple ( $A, B, C$ ) of sets to be minimally intersecting if $|A \cap B|=$ $|B \cap C|=|C \cap A|=1$ and $A \cap B \cap C=\emptyset$. For example, $(\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $N$ be the number of minimally intersecting ordered triples of sets for which each set is a subset of $\{1,2,3,4,5,6,7\}$. Find the remainder when $N$ is divided by 1000. Solution: Logic destroys this problem (it tends to destroy many). We ask ourselves, what conditions must we impose on $A,B,C$ to be considered minimally intersecting? We must have that if $x \in A$, then **either** $x \in B$ or $x \in C$. In particular, we cannot have that $x$ is in all three sets. So to satisfy the conditions of being minimally intersecting, we must choose some $x \in \{1,2,3,4,5,6,7\}$ to be in $A,B$. Call this element $r$ (there are $7$ choices for $r$). Then, from the remaining, we must choose some $x \in \{1,2,3,4,5,6,7\} \backslash \{ r \}$ to be in $B,C$. Call this element $s$ (there are $6$ choices for $s$). Finally, we must choose some $t \in \{ 1,2,3,4,5,6,7 \} \backslash \{ r,s \}$ to be in $C,A$. Once done with this process, we have a "minimally" minimal intersecting triple of sets. For the remaining $4$ elements, we make a choice whether to include it in $A,B,C$ or not in any. Hence, the final answer is $7 \cdot 6 \cdot 5 \cdot 4^4 = 53760$. So, the answer is $760.$