Possible shorter solution to this problem? 
How many pairs of diagonals of a regular decagon are parallel?

The answer is $45$, and is computed via: $$5\binom{4}{2} + 5 \binom{3}{2}$$
which comes as a result of fixing one vertex and choosing $2$ of the $4$ possible parallel diagonals with that vertex fixed, and then fixing a side and choose $2$ of the $3$ possible parallell diagonals with that side fixed.

However, the answer $45$ is also just $\binom{10}{2}$. Is this just a coincidence or is there another, shorter, way to approach this problem?

 A: In a decagon, there are $10\choose2$ ways to pick pairs of vertices that are 1 side apart. Connecting these pairs gives pairs of parallel diagonals. But for each pair, only 5 of the other 9 pairs don't share a share a vertex or result in a diagonal that is just a side of the decagon, so there are $\frac59{10\choose2}$ ways here.
Now we check pairs of vertices where the vertices are 2 sides apart. There are still $10\choose2$ ways to do this where only $\frac39$ don't share vertices, don't give us sides of the decagon, and don't intersect with each other. By intersect, I mean that if you draw a segment that conjoins one pair's points and another for the other pair, then the two don't intersect. Allowing for this would make us double count some pairs of diagonals. So there are $\frac39{10\choose2}$ ways here.
We do the same for pairs of vertices that are 3 sides apart and find that there are $\frac19{10\choose2}$ possibilities
There are none for pairs of vertices that are further apart.
Adding these all up
$$\left(\frac59+\frac39+\frac19\right){10\choose2}={10\choose2}$$
Generalize this to an even n-gon, we have
$$\left(\frac1{n-1}\sum_{i=2}^{n/2-1}n-2i-1\right)\times{n\choose2}$$
