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  "body": "\n## Proof\n\n1. The [[complete graph|03WKB2H474]] $K_n$ has vertex set $V = \\set{1, \\dots, n}$, and any two distinct vertices are adjacent.\n2. Each edge of a [[graph|0JD5C82QMB]] is a $2$-element subset of $V$. By step 1, every $2$-element subset $\\set{u, v}$ of $V$ is an edge of $K_n$.\n3. So the edges of $K_n$ are exactly the $2$-element subsets of $V$. By the [[number of k-element subsets|0BFMR2FQ8B]], there are $\\binom{n}{2}$ of them.\n4. If $n \\ge 2$, the [[factorial formula|033KTKKRV6]] gives $\\binom{n}{2} = \\frac{n!}{2!\\,(n-2)!} = \\frac{n(n-1)}{2}$.\n5. Let $n \\in \\set{0, 1}$. The recursion that defines the [[binomial coefficient|092GEF38YD]] gives $\\binom{0}{1} = \\binom{0}{2} = 0$. Hence $\\binom{1}{2} = \\binom{0}{1} + \\binom{0}{2} = 0$.\n6. For $n \\in \\set{0, 1}$ also $\\frac{n(n-1)}{2} = 0$. So in all cases $K_n$ has $\\binom{n}{2} = \\frac{n(n-1)}{2}$ edges.\n\n## Idea\n\nIn $K_n$ every pair of distinct vertices is an edge, so counting edges is counting pairs. Choosing an ordered pair gives $n(n-1)$ options, and each unordered pair is counted twice.\n",
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