Question

There are $18$ points in a plane such that no three of them are in the same line except five points which are collinear. The number of triangles formed by these points is :

• A. $816$
• B. $806$
• C. $805$
• D. $813$

Answer 1

The correct option is B

$806$

Find the number of triangles formed based on given information

Given, $18$ points five of them are collinear.

We need three non-collinear points to form a triangle.

The number of triangles formed by $18$ points is ${}^{18}{\mathrm{C}}_3$

Since $5$ points are collinear.

Therefore, the number of triangles is ${}^{18}{\mathrm{C}}_3-{}^5{\mathrm{C}}_3$

$$\begin{gathered} =\frac{18!}{3!\left(18-3\right)!}-\frac{5!}{3!\left(5-3\right)!}\left[{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}-\mathrm{r}\right)!}\right] \\ =\frac{18!}{3!\times 15!}-\frac{5!}{3!\times 2!} \\ =\frac{18\times 17\times 16\times 15!}{6\times 15!}-\frac{5\times 4\times 3!}{3!\times 2} \\ =3\times 17\times 16-5\times 2 \\ =816-10 \\ =806 \end{gathered}$$

Hence, option (B) is correct .