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I believe the answer is $\frac12(n-1)^2$, but I couldn't confirm by googling, and I'm not confident in my ability to derive the formula myself.

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    $\begingroup$ So you think a clique of $2$ vertices has exactly $\frac12$ edges? $\endgroup$ Commented Sep 20, 2012 at 13:20
  • $\begingroup$ Oh yeah, how silly. Good thing I checked. $\endgroup$
    – MikeFHay
    Commented Sep 20, 2012 at 13:21
  • $\begingroup$ $\frac12 (n^2 - n)$? $\endgroup$
    – MikeFHay
    Commented Sep 20, 2012 at 13:25
  • $\begingroup$ I think I would upvote this question if it included your (as Chris points out, incorrect) derivation. $\endgroup$ Commented Sep 20, 2012 at 13:26
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    $\begingroup$ @MikeL: I think I would upvote this question if you made much effort, then :P $\endgroup$ Commented Sep 20, 2012 at 13:37

2 Answers 2

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A clique has an edge for each pair of vertices, so there is one edge for each choice of two vertices from the $n$. So the number of edges is:

$$\binom{n}{2}=\frac{n!}{2!\times(n-2)!}=\frac{1}{2}n(n-1)$$

Edit: Inspired by Belgi, I'll give a third way of counting this! Each vertex is connected to $n-1$ other vertices, which gives $n(n-1)$ times that an edge is joined to a vertex. As each edge is joined to exactly two vertices, there must be $\frac{1}{2}n(n-1)$ edges.

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  • $\begingroup$ That makes sense, thank you. $\endgroup$
    – MikeFHay
    Commented Sep 20, 2012 at 13:31
  • $\begingroup$ nice, I wish I could upvote again :) $\endgroup$
    – Belgi
    Commented Sep 20, 2012 at 13:49
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Another way to calculate what Matt said it to this: number the vertices from $1$ to $n$, and consider the graph with $n$ vertices but with no edges.

Take the first vertice: it has vertics to the other $n-1$ vertices, connect those $n-1$ vertices

Take the second vertice: it has vertics to the other $n-1$ vertices - but one of them is already connected - connect the other $n-2$

do this untill the last vertice. you get that you connected $(n-1)+(n-2)+...+1$ vertices which is an arithmetic proggrestion whose sum is $\frac{1}{2}(n-1)n$

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  • $\begingroup$ Yes, I think this is how I once learned it. Thank you. $\endgroup$
    – MikeFHay
    Commented Sep 20, 2012 at 13:33

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