Ten [points](https://artofproblemsolving.com/wiki/index.php/Point) in the plane are given, with no three [collinear](https://artofproblemsolving.com/wiki/index.php/Collinear).  Four distinct [segments](https://artofproblemsolving.com/wiki/index.php/Segment) joining pairs of these points are chosen at random, all such segments being equally likely.  The [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that some three of the segments form a [triangle](https://artofproblemsolving.com/wiki/index.php/Triangle) whose vertices are among the ten given points is $m/n,$ where $m_{}$ and $n_{}$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime) [positive](https://artofproblemsolving.com/wiki/index.php/Positive) [integers](https://artofproblemsolving.com/wiki/index.php/Integer).  Find $m+n.$
First, let us find the number of triangles that can be formed from the 10 points. Since none of the points are collinear, it is possible to pick ${10\choose3}$ sets of 3 points which form triangles. However, a fourth distinct segment must also be picked. Since the triangle accounts for 3 segments, there are $45 - 3 = 42$ segments remaining.
The total number of ways of picking four distinct segments is ${45\choose4}$. Thus, the requested probability is $\frac{{10\choose3} \cdot 42}{{45\choose4}} = \frac{10 \cdot 9 \cdot 8 \cdot 42 \cdot 4!}{45 \cdot 44 \cdot 43 \cdot 42 \cdot 3!} = \frac{16}{473}$. The solution is $m + n = 489$.