How many diagonals can be drawn in the pentagon?

There are 5 diagonals, no repeats. For example, AD and DA are the same diagonal, so it should be a combination.

I think the number of diagonals that can be drawn in any poygon with n vertexes (sides), where n >3, is given by: (n)(n-3)/2.

To see this, note that no diagonal can be drawn from a vertex to itself, nor can any diagonal be drawn to the two neighboring vetexes. So this leaves (n-3) vertexes left. And from each vertex, the same number of diagonals can be drawn. So the total number of diagonals that can be drawn is (n)(n-3). But the diagonal from v_i to v_j is the same diagonal drawn from v_j to v_i, so we must divide this total number by two to prevent "double counting."

So we have...... (n)(n-3) / 2.

Thank you Kitty, that is a good answer.

However, I would also like to see this done using combination notation.

Thank you Chris, you must have been reading my mind. 

P. S.........there may be some combinatoric that could be found for this, but I couldn't think of any other way to do it!! Anyone else have any ideas??

n=3: the number of diagonals = 0 + 1 + 2   -  3

n=4: the number of diagonals = 0 + 1 + 2 + 3  -  4

n=5: the number of diagonals = 0 + 1 + 2 + 3 + 4  -  5

n=6: the number of diagonals = 0 + 1 + 2 + 3 + 4 + 5 -  6

...

n: the number of diagonals = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + ... + (n-1)  -  n

the number of diagonals = $$\frac{(1+(n-1))*(\;((n-1)-1)\;+\;1\;)}{2} - n = \frac{n(n-1)}{2} - n=\binom{n}{2}-n$$

the number of diagonals = $$\boxed{\binom{n}{2}-n=\frac{n(n-3)}{2}}\quad n\ge3$$

S.

Thank you anonymous.  This has been a really interesting question.

Thank you to Chris and Kitty as well!