Question

There are $p$ points in a plane, no three of which are in the same straight line with the exception of $q$, which are all in the same straight line; find the number (1) of straight lines, (2) of triangles which result from joining them.

Answer 1

$\left(1\right)$ There are two points required for a single straight line.
$\therefore$ Total number of single straight lines formed by ${}^{\prime }{p}^{\prime }$ points $={}^p{C}_2$
But ${}^{\prime }{q}^{\prime }$ points are already in a straight line i.e. ${}^q{C}_2$
$\therefore$ Total number of straight lines $={}^p{C}_2-{}^q{C}_2+1$ (q points form a straight line)
$=\frac{p(p-1)}{2}-\frac{q(q-1)}{2}+1$
$\left(2\right)$ There are $3$ points required for triangle making
No. of triangles formed out of ${}^{\prime }{p}^{\prime }$ points $={}^p{C}_3$
No. of triangles formed out of ${}^{\prime }{q}^{\prime }$ points $={}^q{C}_3$
Since ${}^{\prime }{q}^{\prime }$ points are in a straight line.
Hence, they cannot form a triangle.
$\therefore$ Number of triangles formed $={}^p{C}_3-{}^q{C}_3$
$=\frac{p(p-1)(p-2)}{6}-\frac{q(q-1)(q-2)}{6}$