Question

The sides $AB,BC,CA$ of a triangle $ABC$ have $3,4$ and $5$ interior points respectively on them. The number of triangles that can be constructed using these points as vertices is

• A. $205$
• B. $210$
• C. $315$
• D. $216$

Answer 1

The correct option is A $205$

Total number of points $12$. If no three points are co-linear then total number of the triangles would be ${}^{12}{C}_3$.

But $3$ points on AB , $4$ points on BC and $5$ points on CA are co-linear.

So, total number of triangles formed should be

$\begin{array}{c}{=}^{12}{C}_3-\left({}^3{C}_3{+}^4{C}_3{+}^5{C}_3\right)\\ =220-\left(1+4+10\right)\\ =205\end{array}$