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 pointsfrom
AB )
or (one point from BC and two points from AC)
or(one point from
BC and two points from AC)
or(one point from
AC and two points from BC)
⇒ Total number of triangles
r=(3C1×5C1×6C1)+(3C1×5C2)+(3C1×6C2)+(6C1×3C2)+(5C2)+(6C2)=90+30+15+45+18+75+60=333[ using nCr= and n!=1×2×3×...×n] Method (II) (Direct Method)
Total number of points
=3+5+6=14Then, when we construct a triangle, we must select 3 points out of 14 but these points never be collinear.
∴ Total number of triangles formed
=14C3−3C3−5C3−6C3=333