Question

There are $10$ points in a plane of which $4$ are collinear. The number of quadrilaterals that can be formed is

• A. $15$
• B. $45$
• C. $50$
• D. $185$

Answer 1

Answer: (D)

To form a quadrilateral, $4$ points are required. They can be
(i) All the $4$ points from $10-4=6$ non-collinear points.
(ii) $3$ points from $6$ non-collinear points and $1$ point from $4$ collinear points.
(iii) $2$ points from $6$ non-collinear points and $2$ points from $4$ collinear points.
$\therefore$ reqd. no. of ways $={}^6{C}_4+{}^6{C}_3\cdot {}^4{C}_1+{}^6{C}_2\cdot {}^4{C}_2$
$=\frac{6\times 5}{1\times 2}+\frac{6\times 5\times 4}{1\times 2\times 3}\times 4+\frac{6\times 5}{1\times 2}\cdot \frac{4\times 3}{1\times 2}$
$=15+80+90$
$=185$