Counting the number of connected undirected simple graphs with \(n\) distinct vertices and \(k\) edges.
|dw:1429477030300:dw| \[ n-1<k<{n\choose 2} \]
\[ \begin{array}{c|c|c} n&k_{\min}&k_{\max} \\ \hline 2&1&1\\ 3&2&3\\ 4&3&16\\ 5&4&10\\ 6&5&15\\ 7&6&21\\ 8&7&28 \end{array} \]
\[ n-1\leq k\leq {n\choose 2} \]
\[ {n\choose 2} = \frac{n(n-1)}{2} \]
|dw:1429479396090:dw| this a graph with 2 vertices and 0 edges so here we have n=2 and k=0 but your inequality says this \[n-1\leq k\leq {n\choose 2}\] \[1 \leq 0 \leq 0\] but 0 isn't greater than 1
Are we only looking at what its called? connected graphs or is it called complete graphs? can't remember
Connected graphs
I think connected has something to do with directions and complete graphs is actually what you are looking for https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=13&cad=rja&uact=8&ved=0CGcQFjAM&url=https%3A%2F%2Fwww.math.ku.edu%2F~jmartin%2Fcourses%2Fmath105-F11%2FLectures%2Fchapter6-part2.pdf&ei=fSE0VYyqBseRoQSskoHIAw&usg=AFQjCNFip_uWUDFQT4wcX5dk497V09RYvQ&bvm=bv.91071109,d.cGU
|dw:1429479395167:dw| so here's the adjacency matrix for this graph so reading down the second column this means 1 connects to 2, 2 doesn't connect to itself, and 2 doesn't connect to 3. \[\left[ \begin{array}c 0 & 1 & 1\\1 & 0 & 0\\1 & 0 & 0\\\end{array} \right]\] Squaring this matrix gives us \[\left[ \begin{array}c 2 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\\end{array} \right]\] which seems weird, but makes sense especially in the context of directed graphs. Just imagine the edges are all directed in both directions.
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