# Number of triangles formed by all chords between $n$ points on a circle

We have $n$ point on circumference of a circle. We draw all chords between this points. No three chords are concurrent.

How many triangles exist that their apexes could be on circumference of circle or intersection points of chords and side of triangles is on chords?

Things I have done so far:

There are four situations:

• situation #1:all points on circumference.
• situation #2:two points on circumference,one intersection point.
• situation #3:one points on circumference,two intersection point.
• situation #4:three intersection point.

Situation #1 is easily countable. ${n \choose 3}$

For other situations, I can't find a way for counting them.

Answer (according to answer key): $${n \choose 3} + 4{n \choose 4} + 5{n \choose 5} + {n \choose 6}$$

• possible duplicate of How many triangles are formed by $n$ chords of a circle? – Saharsh Jul 27 '14 at 6:41
• – Saharsh Jul 27 '14 at 6:43
• @Alpha,the OP in this question math.stackexchange.com/questions/313489 don't have problems in figuring out why number of triangles are ${n \choose 3}$+$4{n \choose 4}$+$5{n \choose 5}$+${n \choose 6}$.and the second one you posted is not about combinations. – user2838619 Jul 27 '14 at 6:45

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.