Question

The sides $\mathrm{AB},\mathrm{BC},\mathrm{CA}$ of a triangle $\mathrm{ABC}$ have respectively $3,4$ and $5$ points lying on them.

The number of triangles that can be constructed using these points as vertices is?

• A. $205$
• B. $220$
• C. $210$
• D. None of these

Answer 1

The correct option is A

$205$

Explanation for correct option:

Given that the sides $\mathrm{AB},\mathrm{BC},\mathrm{CA}$ of a triangle $\mathrm{ABC}$ have respectively $3,4$ and $5$ points lying on them.

Total number of points$=3+4+5$

$=12$

Number of triangles that can be formed with these $12$ points $={}^{12}\mathrm{C}_3$

Since, we have $3$ points on $\mathrm{AB}$, $4$ points on $\mathrm{BC}$ and $5$ points on $\mathrm{CA}$.

Therefore, number of triangles that can be formed $={}^{12}\mathrm{C}_3-\left({}^3\mathrm{C}_3+{}^4\mathrm{C}_3+{}^5\mathrm{C}_3\right)$

$=\frac{12!}{3!9!}-\left(\frac{3!}{3!0!}+\frac{4!}{3!1!}+\frac{5!}{3!2!}\right)$ $\left[\because {}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}-\mathrm{r}\right)!}\right]$

$=\frac{12·11·10}{3.2}-\left(\frac{1}{1}+\frac{4}{1}+\frac{5.4}{2}\right)$

$=220-\left(1+4+10\right)$

$=205$

Hence, the correct option is option(A) i.e. $205$