To form a pair of chords that intersect, you need 4 points. Hence, that can be done in nC4 ways. There is only way to join the 4 points with two chords and form an intersection. Hence, the final answer is nC4.
Situation $2$:
For each selection of any $4$ points on the circumference, you can draw the diagram you have. The line segment that joins the adjacent circumference points, could instead join any of the $4$ pairs of adjacent circumference points, so we have $4$ different triangles for each choice of $4$ circumference points.
There are $\binom{n}{4}$ ways to choose the $4$ points, so Situation $2$ contributes:
$\qquad4 \binom{n}{4}$ triangles.
Situation $3$:
For each selection of any $5$ points on the circumference, you can draw the diagram you have. You could choose any of these $5$ points to be a vertex of a Situation $3$ triangle, so we have $5$ different triangles for each choice of $5$ circumference points.
There are $\binom{n}{5}$ ways to choose the $5$ points, so Situation $3$ contributes:
$\qquad5 \binom{n}{5}$ triangles.
Situation $4$:
For each selection of any $6$ points on the circumference, you can draw the diagram you have. There is only one way to construct that internal triangle given these $6$ circumference points.
There are $\binom{n}{6}$ ways to choose the $6$ points, so Situation $4$ contributes:
$\qquad \binom{n}{6}$ triangles.
Best Answer
Suppose you have three concurrent lines (i.e. all intersecting at a single point). Now perturb the position of the line endpoints slightly. Almost surely the lines no longer intersect a single point; instead you have three pairwise intersections and a tiny triangle where the single intersection point used to be. Therefore any time you have (exactly) three concurrent lines, you "lose" two intersection points from the number predicted by your formula.