How many triangles can be formed by joining 6 points lying on a circle?

Hint: Here we know that we need exactly 3 points to construct a triangle. So here we have to use a combination concept to solve this problem. And we also know that none of the points should be collinear to form a triangle.

Complete Step-by-Step solution:
Given that as all the 6 points lie on the circle none of them are collinear.
Now we know that to make a triangle we need 3 points. Hence we have picked 3 points of 6 given points.
Then we know that the number of different ways to pick 3 points out of 6 given points is
${ \Rightarrow ^6}{C_3}$
We know that $^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
So, we can write$^6{C_3} = \dfrac{{6!}}{{(6 - 3)!3!}}$
$ \Rightarrow \dfrac{{6!}}{{3!3!}}$
$ \Rightarrow \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1 \times 3 \times 2 \times 1}}$
=20
Therefore 20 triangles can be formed by joining 6 points lying on a circle.

NOTE: Here we have to remember that collinear points cannot form triangles. Since the given points lie in the circle then we can say that none of the points are collinear which means all the 6 points are distinct.