How many triangles, can be formed?

Question

I have an excersie that says:

¿How many triangles can be formed, with the vertex of a hexagon?

I think this:

$1$) Ok, it is requesting triangles, therefore, it is not permutation since it takes sets of 3 vertex.

Now it should be variation or combination, then I thought:

¿The triangle formed by the verte $ ABC $ is equal to $ BCA $? Well yes, since that combination of vertex has already been used.

So, now I must know if it is combination with or without repetition

The vertex can be repeated, for example:

$ ABC $ is equivalent to $ CBA $

But,

$ ABC $ is different from $ CBF $, since $ BC $ is repeated but $ F $ is not repeated, so are different triangles.

Ok, I use the formula: $\frac{(n +r - 1)!}{(n - 1)! * r!}$

$n = 6$ ( number of hexagon vertex ), $r = 3$ ( number of triangle vertex)

$= \frac{8!}{5! * 3!} = \frac{336}{!3} = 56$

But the correct is wrong, and the correct result should be $20$.

¿What is my error?

PD:

I used the formula for repetition, and all answer used the other formula, why ?

Answer 1

It's just $C_3^6=20$. (choosing 3 points from 6).

Answer 2

You should just be selecting three vertices out of six, and ${6 \choose 3}=\frac {6!}{3!(6-3)!}=\frac {720}{6\cdot 6}=20$