Given the number of vertices in a graph, does the formula for max number of edges it can have make intutive sense? I hope I'm able to put the question clearly.
Suppose a pen costs $x$ dollars, and I buy 3 pens. So the formula for total cost is: $3x$
This makes sense, since total cost is cost of one pen times number of pens.
For a graph with n vertices, there are at the most $1+2+3....(n-1)$ edges it can have. Which is basically $(n-1)n/2$
How can I make similar intutive sense out of this formula?
 A: Consider the following table. The top row and lefthand column list the $n$ vertices of a graph. The body of the table shows an $X$ or an $O$ for each possible edge. Notice, though that each possible edge is shown twice, once marked $X$ and once marked $O$. Specifically, if $i<k$, the possible edge between $v_i$ and $v_k$ is marked $X$ in row $i$, column $k$, and $O$ in row $k$, column $i$. Thus, to count the number of possible edges, we should count either the $X$’s or the $O$’s, but not both.
$$\begin{array}{r|c|c}
&v_1&v_2&v_3&\dots&v_{n-1}&v_n\\ \hline
v_1&-&X&X&\dots&X&X\\ \hline
v_2&O&-&X&\dots&X&X\\ \hline
v_3&O&O&-&\dots&X&X\\ \hline
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ \hline
v_{n-1}&O&O&O&\dots&-&X\\ \hline
v_n&O&O&O&\dots&O&-
\end{array}$$
As you can see from the arrangement, the number of $X$’s (or $O$’s) is clearly $$1+2+3+\ldots+(n-1)\;.$$ On the other hand, there are $n^2$ cells in the table, of which $n$ are on the diagonal and represent impossible edges, so there are $n^2-n$ $X$’s and $O$’s altogether. We want just the $X$’s (or just the $O$’s), so we want half of that number, which is $$\frac{n^2-n}2=\frac{n(n-1)}2\;.$$ We conclude, then, that the number of possible edges is 
$$1+2+3+\ldots+(n-1)=\frac{n(n-1)}2\;.$$
If you know about binomial coefficients, you can see this in another way: each edge corresponds to one possible choice of two of the $n$ vertices, and there are $$\binom{n}2=\frac{n(n-1)}2$$ ways of choosing two from a set of $n$ objects.
A: You can also argue by induction: let us have a graph with $n$ vertices and max. number of edges, then consider an (n+1)th vertex: you can draw $+n$ edges.
For one vertex there cannot be edges, and hence it is
$0+1+2+\ldots+(n-1)$ for $n$ edges.
By the way, one could also consider graphs with loops or parallel arrows, even directed graphs.. If multiple arrows allowed, then there is no upper limit on their numbers,
A: Cut each edge in the middle, making two half-edges, one dangling from each endpoint. Now, each vertex can have a maximum of $n-1$ half-edges, one reaching towards each of the other vertices. $n$ vertices, each with $n-1$ half-edges attached to it, makes a total of $n(n-1)$ half-edges, or $\frac{n(n-1)}2$ whole edges.
A: Note: This problem is intuitively the same as the handshake problem where instead of counting the max handshakes among $n$ people you want to find the max edges among $n$ vertices.

Suppose we have $n$ vertices labeled $v_i$,  $(i = 1,...,n)$ (the actual order is irrelevant) :
Step 1) Take vertex $v_1$ and connect it to all other $(n-1)$ vertices, now disgard $v_1$.
Step 2) Take vertex $v_2$ and connect it to all other $(n-2)$ vertices, now disgard $v_2$.
$$...$$
Step n-2) Take $v_{n-2}$ and connect it to all other $(2)$ vertices, now disgard  $v_{n-2}$.
Step n-1) Take $v_{n-1}$ and connect it to all other $(1)$ vertices, now disgard  $v_{n-1}$.
Now, notice the number of edges (in parenthesis) above we add in each step can be written as:$$\sum\limits_{i=1}^{n-1} i = \frac{n(n-1)}{2}$$
