To prove that the number of edges in a complete graph $K_n$ with $n$ vertices is given by $\frac{n(n-1)}{2}$, let's proceed by analyzing how edges are formed in a complete graph.

Step-by-Step Proof

1. Definition of a Complete Graph: A complete graph $K_n$ is a simple, undirected graph where every pair of distinct vertices is connected by a unique edge.

2. Counting Pairs of Vertices: Each edge in $K_n$ connects a unique pair of vertices. Since there are $n$ vertices in total, we need to determine how many ways we can select pairs of vertices from these $n$ vertices.

3. Using Combinations to Count Pairs: The number of ways to choose 2 vertices from $n$ vertices (to form an edge) is given by the combination formula: $\binom{n}{2} = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2}$

4. Conclusion: Each pair of vertices corresponds to a unique edge in $K_n$, so the total number of edges in $K_n$ is precisely $\binom{n}{2}$, which simplifies to: $\frac{n(n-1)}{2}$

Therefore, the number of edges in a complete graph with $n$ vertices is $\frac{n(n-1)}{2}$, as required.

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Would you like a further breakdown of any step or a visual example? Here are some related questions to explore this concept:

1. How does the formula for the number of edges change if the graph is not complete?
2. How can we generalize this result for directed complete graphs?
3. What is the relationship between the degree of each vertex and the number of edges in $K_n$?
4. How does the proof change if we consider multiple edges between vertices?
5. Can this combinatorial approach apply to other graph structures, like bipartite graphs?

Tip: In combinatorics, breaking down problems by counting distinct pairs (using combinations) is a powerful technique to solve similar problems involving relationships among items.