Bermuda Triangle

Firstly we must approach the case when two points are chosen on one side and two points are chosen from another, leaving one side with no selected points. This can be determined through the use of combinations and summing the various possibilities. Below is a list of the different scenarios.

2 points are chosen on side $PR$ and connected to 2 points on side $PQ$ .

2 points are chosen on side $PR$ and connected to 2 points on side $QR$ .

2 points are chosen on side $PQ$ andf connected to 2 points on side $QR$

These scenarios are calculated and then summed below.

$\binom{6}{2}\binom{4}{2} + \binom{6}{2}\binom{3}{2} + \binom{4}{2}\binom{3}{2} = 153$

Now we must approach the second case when two points are chosen from one side, one point from a second side and the fourth point from the last side. Below are the 3 possible scenarios for this case.

2 points are chosen on side $PR$ and connected to 1 point on $PQ$ and 1 point on $QR$ .

1 point is chosen on side $PR$ and connected to 2 points on $PQ$ and 1 point on $QR$ .

1 points is chosen on side $PR$ and connected to 1 point on $PQ$ and 2 points on $QR$ .

The number of intersections from these scenarios are calculated and summed below.

$\binom{6}{2}\binom{4}{1}\binom{3}{1} + \binom{6}{1}\binom{4}{2}\binom{3}{1} + \binom{6}{1}\binom{4}{1}\binom{3}{2} = 360$

Adding the number of intersections from the two cases yields the total number of intersections.

$153 + 360 = 513$

$\therefore$ There are 513 different points of intersection within the triangle $PQR$ when all possible lines are drawn.