How many ways to create and select 2 lines from a set of 5 points We have 5 points: A, B, C, D, E
Any two unique points define a line.
How many ways can we select two unique lines?
example) AB-AC form a set of 2 lines. AB-AD, AB-AE, AB-BC, AB-BD, etc.
I tried brute force counting, and I got 40.
However, I was expecting that answer to match a calculated answer of:
$$L={{5}\choose{2}} $$
where $L$ is the number of possible lines.
$$N=.{{L}\choose{2}}$$
where $N$ is the possible ways we can select a pair of lines that can be defined by the 5 points.
But, a quick evaluation gives 45 instead of 40. Possible that I messed up somewhere along the brute force method, but I want to verify that the equations above are correct for solving this problem so that I can generalize this to a larger set of points. 
Thanks!
EDIT: No 3 points are co-linear
EDIT: Here are the 40 combos that I found. I have not found my error yet, but would like to so i can fix my for loop
1:2 - 1:3
1:2 - 1:4
1:2 - 1:5
1:2 - 2:3
1:2 - 2:4
1:2 - 2:5
1:2 - 3:4
1:2 - 3:5
1:2 - 4:5
1:3 - 1:4
1:3 - 1:5
1:3 - 2:3
1:3 - 2:4
1:3 - 2:5
1:3 - 3:4
1:3 - 3:5
1:3 - 4:5
1:4 - 1:5
1:4 - 2:4
1:4 - 2:5
1:4 - 3:4
1:4 - 3:5
1:4 - 4:5
1:5 - 2:5
1:5 - 3:5
1:5 - 4:5
2:3 - 2:4
2:3 - 2:5
2:3 - 3:4
2:3 - 3:5
2:3 - 4:5
2:4 - 2:5
2:4 - 3:4
2:4 - 3:5
2:4 - 4:5
2:5 - 3:5
2:5 - 4:5
3:4 - 3:5
3:4 - 4:5
3:5 - 4:5

Yes, your calculation is correct: There are 
$${5\choose 2} = 10$$ lines, and so there are 
$${10\choose 2} = 45$$ pairs of lines.
So you must have missed some when doing this by brute force.
And yes, this clearly generalizes for $n$ points:
$${n\choose 2} = \frac{n(n-1)}{2}$$ lines  and thus
$${{\frac{n(n-1)}{2}}\choose 2} = \frac{\frac{n(n-1)}{2}\cdot(\frac{n(n-1)}{2}-1)}{2}= \frac{n^2(n-1)^2}{8} - \frac{n(n-1)}{4}$$ pairs
