Question

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

• A. $220$
• B. $205$
• C. $190$
• D. $85$

Answer 1

Answer: (B)

$205$

Total number of points $12.$ If no three points are co-linear$,$

Th en$,$ total number of 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$:$

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

Hence,

option $\left(B\right)$ is correct answer.