Question

Could anyone explain to me how formula of finding total number of triangles formed by joining the vertices of n-sided polygon is derived?

Answer 1

How is this formula derived?\[N = \frac{ n(n-1)(n-2) }{ 6 }\]
Source: http://math.stackexchange.com/questions/1446438/how-many-triangles-can-be-formed-by-the-vertices-of-a-regular-polygon-of-n-sides

Answer 2

SInce number of verticles is equal to \[\frac{ n(n-3) }{ 2 }\] I would think that number of ways to chose one of them would be: \[\frac{ n(n-3) }{ 2 }\cdot (\frac{ n(n-3) }{ 2 }-1) \cdot (\frac{ n(n-3) }{ 2 }-2)\]

Answer 3

I am very sorry, that I had bothered you. I understood that my interpretation of that answer has been wrong. Very sorry!

Answer 4

n choose 3 (combination) nC3 is defined as \(\frac{n!}{(n-3)!3!}=\frac{n(n-1)(n-2)}{3!}=\frac{n(n-1)(n-2)}{6}\).

Answer 5

This number includes only triangles that are formed by vertices, not those formed by intersection of the diagonals.