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Home » Die Fläche ist \( 6 \cdot 12 = 72 \) Quadratmeter. - Baxtercollege

Die Fläche ist \( 6 \cdot 12 = 72 \) Quadratmeter. - Baxtercollege

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 So for each vertex \( v \), the number of paths of length 2 with center \( v \) is \( inom{\deg(v)}{2} \), and each such path \( u-v-w \) uses edges \( uv \) and \( vw \), both in \( E \). Then the trio \( \{u,v,w\} \) has exactly two close pairs (if no other edge), and since \( u \) and \( w \) are not necessarily connected, itâs exactly 2.

📌 But we must ensure that no additional edge exists, but since we are counting based on edge set, and the trio is defined by the three vertices, and the condition is on the number of close pairs among the three edges, if two are in \( E \) and one is not, then the trio has exactly two close pairs.

📌 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 a triple \( \{u,v,w} \) to have exactly two close pairs, it must be that two of the three pairs are in \( E \), and the third is not.

📌 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 \( E \).

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