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- So total number of such trios is the number of unordered triples \( \{u,v,w\} \) such that exactly two of the three pairs \( (u,v), (v,w), (u,w) \) are in \( E \).
- Each such triple is counted once, and corresponds to a middle vertex \( v \) with neighbors \( u,w \), and both edges present.
- So sum over \( v \), number of ways to choose two neighbors: \( inom{\deg(v)}{2} \), but only if all three pairs are not both present â but we are counting triples where *exactly* two pairs are in \( E \), so if the third pair \( (u,w) \) is also in \( E \), then the trio has three close pairs â not allowed.
- So for vertex \( v \), number of pairs \( (u,w) \) such that \( uv \in E \), \( vw \in E \), but \( uw
- otin E \), and \( \{u,w\} \) is not in \( E \).
- So we must count, for each \( v \), the number of unordered pairs \( \{u,w\} \subseteq N(v) \) such that \( uw
