Let's call the $n$ points on the circle as circle-points and those intersection points as inner-points. Then triangles formed by the circle-points and inner-points can be grouped into the following four categories:

• $\mathbb{A}$: all vertices are circle-points
• $\mathbb{B}$: two vertices are circle-points and one is inner-point
• $\mathbb{C}$: one vertex is circle-point and two are inner-points
• $\mathbb{D}$: all vertices are inner-points

Every triangle in $\mathbb{A}$ has a bijective relation with three distinct points on the circle. Hence, $\mid\mathbb{A}\mid=\binom{n}{3}$ where $\mid\mathbb{S}\mid$ denotes the number of elements in set $\mathbb{S}$.

For $\mathbb{B}$, let's consider any inner-point $O$ as shown in the first diagram below. An inner point $O$ determines four triangles in $\mathbb{B}$. Meanwhile, $O$ has a bijective relation with four circle-points. This means that these five points have a bijective relation with four triangles in $\mathbb{B}$: $\triangle{A_1A_2O}$, $\triangle{A_2A_3O}$, $\triangle{A_3A_4O}$ and $\triangle{A_4A_1O}$. Hence, $\mid\mathbb{B}\mid=4\binom{n}{4}$.

For $\mathbb{C}$, as shown in the right graph above, every $5$ circle-points have a bijective relation with $5$ inner-points. All these $10$ points have a bijective relation with $5$ triangles in $\mathbb{C}$. Hence, $\mid\mathbb{C}\mid=5\binom{n}{5}$.

For the last case, there are $\mid\mathbb{D}\mid=\binom{n}{6}$ such triangles. This is because given any six points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ in that order but not necessarily next to each other, the intersection points created by segments $A_1A_4$, $A_2A_5$ and $A_3A_6$ has a one-to-one mapping with a triangle in $\mathbb{D}$.

Therefore, the final answer is $$\boxed{\binom{n}{3}+4\binom{n}{4}+5\binom{n}{5}+\binom{n}{6}}$$