Let’s solve the problem step by step: ### Step 1: Understand the Condition **Condition:** For every non-empty subset \( W \subseteq V \), the subgraph induced by \( W \) is connected. **Explanation:** This means that **no matter which vertices you pick**, the subgraph formed by just those vertices (and all the edges between them in G) is connected. In particular, for any pair of vertices, the induced subgraph on just those two vertices is connected. That is only possible if there is an edge between **every** pair of vertices. --- ### Step 2: Characterize the Graph **Result:** The only graphs where every induced subgraph is connected are **complete graphs**. **Explanation:** - If G is not complete, then there exists at least one pair of vertices \( u, v \in V \) such that there is **no edge** between them. - Consider the subset \( W = \{u, v\} \). The induced subgraph is just two isolated vertices, which is **not connected**. - Thus, G must be a **complete graph** on n vertices, denoted \( K_n \). --- ### Step 3: Count the Number of Edges in a Complete Graph **Result:** A complete graph on \( n \) vertices has \( \binom{n}{2} = \frac{n(n-1)}{2} \) edges. **Explanation:** - In \( K_n \), every pair of vertices has an edge between them. - There are \( n \) vertices, and each pair forms a unique edge. - The total number of unordered pairs of \( n \) vertices is \( \binom{n}{2} \). --- ## Final Answer **The number of edges in \( G \) is** \[ \boxed{\frac{n(n-1)}{2}} \] where \( n = |V| \). **Proof Summary:** If any induced subgraph is to be connected, every pair of vertices must be adjacent—so G must be complete, and the number of edges in \( K_n \) is \( \frac{n(n-1)}{2} \).

Let's approach this problem systematically in two steps, providing explanations at each stage and concluding with the final answer. --- ### Step 1: Understand the given condition and its implications **Condition:** For *every non-empty subset* \( W \subseteq V \), the subgraph induced by \( W \) is connected. **Explanation:** - This is a very strong condition. It means that no matter which subset of vertices you pick, the subgraph they form must be connected. - In particular, consider subsets of size 2 (pairs of vertices). For any two vertices \( u, v \), the induced subgraph on \( \{u, v\} \) must be connected, which can only happen if there is an edge between them. **Implication:** - Since such a condition holds for all pairs, **every pair of vertices must be connected by an edge**. - Therefore, the graph must be **complete**. --- ### Step 2: Determine the number of edges in this complete graph **Result:** - The graph \( G \) must be the complete graph \( K_n \) on \( n \) vertices. **Number of edges in \( K_n \):** \[ \boxed{\frac{n(n-1)}{2}} \] **Explanation:** - In a complete graph, every pair of distinct vertices has an edge. - The total number of such pairs (and thus edges) is the number of 2-element subsets of an \( n \)-element set, which is \( \binom{n}{2} = \frac{n(n-1)}{2} \). --- ## Final Answer: \[ \boxed{\frac{n(n-1)}{2}} \] **Summary:** Since the induced subgraph on any subset must be connected, the only way is for the graph to be complete, and a complete graph on \( n \) vertices has \( \frac{n(n-1)}{2} \) edges.