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)

$185$

Answer 2

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 $=\,^{6}C_{4}+\,^{6}C_{3}\cdot\,^{4}C_{1}+\,^{6}C_{2}\cdot\,^{4}C_{2}$ $=\frac{6\times5}{1\times2}+\frac{6\times5\times4}{1\times2\times3}\times4+\frac{6\times5}{1\times2}\cdot\frac{4\times3}{1\times2}$ $= 15 + 80 + 90$ $=185$