Thank you for reporting, we will resolve it shortly
Q.
There are $10$ points in a plane of which $4$ are collinear. The number of quadrilaterals that can be formed is
Permutations and Combinations
Solution:
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 $=\,{}^{6}C_{4}+\,{}^{6}C_{3}\cdot\,{}^{4}C_{1}+\,{}^{6}C_{2}\cdot\,{}^{4}C_{2}$
$=\frac{6\times5}{1\times2}+\frac{6\times5\times4}{1\times2\times3}\times4+\frac{6\times5}{1\times2}\cdot\frac{4\times3}{1\times2}$
$= 15 + 80 + 90$
$=185$