Question

$110$ triangles can be formed by joining $10$ points as vertices, in which $n$ points are collinear. Then the value of $n$ is

• A. $5$
• B. $6$
• C. $3$
• D. $4$

Answer 1

The correct option is A $5$

So the number of triangles formed,
${}^{10}{C}_3{-}^n{C}_3=110$
$\Rightarrow \frac{10\times 9\times 8}{6}{-}^n{C}_3=110$
$\Rightarrow 120-110{=}^n{C}_3$
$\Rightarrow 10{=}^n{C}_3$
$\Rightarrow 10=\frac{n!}{(n-3)!3!}$
$\Rightarrow 10=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!6}$
$\Rightarrow 60=n(n-1)(n-2)$
$\therefore n=5$