Permutations and Combinations Part 1

Method (I) (Proper Method)
Whenever we construct a triangle, we must require three non-collinear points.∴ Total number of triangles using the points 3,5 and 6 which are on the sides AB,BC and CA
= Either taking (one point from AB,BC and CA ) or (one point from AB and two points from BC )or
(one point from BC and two points from AB ) or
(one point from AB and two points from AC ) or
(one point from AC and two points from AB ) or
(one point from BC and two points from AC ) or
(one point from AC and two points from BC )
⇒ Total number of triangles
=90+30+15+45+18+75+60
=333
(using ‌nCr=‌

n!

r!(n−r)!

and n!=1×2×3×...×n