Question

There are $12$ points in a plane. The number of straight lines joining any two of them, when $3$ of them are collinear, is

• A. $60$
• B. $63$
• C. $64$
• D. $65$

Answer 1

Answer: (B)

$63$

Answer 2

Required number ways ${{=}^{12}}{{C}_{2}}{{-}^{3}}{{C}_{1}}$
$=\frac{12\times 11}{1\times 2}-3$
$=6\times 11-3$
$=66-3=63$