Question

There are $20$ points in the plane of which $5$ are collinear. Find the maximum number of distinct (a) straight lines (b) triangles (c) quadrilaterals that can be formed by joining these points.

Answer 1

Hint:
Straight lines require only two point minimum and the triangles require the minimum of the three points to be drawn and the quadrilateral require minimum of the four points. We’ll use this information to count the total points.

Complete step by step solution:
(a) So here we have $20$points in the plane of which $5$ are collinear. Here collinear means that all the five points must lie on the same line. So we can say that we can get only one straight line drawn from the five collinear points and from each of the five pints there will be another $15$ lines joining to other $15$ points. So the number of the lines will be $5(15) = 75$
Now those $15$ points will have lines among themselves which would be ${}^{15}{C_2}$ which means that we need at least two different points to draw the straight line so the number of straight lines will be
$
   = 1 + 75 + {}^{15}{C_2} \\
   = 76 + \dfrac{{15!}}{{13!2!}} = = 76 + 105 = 181{\text{ straight lines}} \\
 $
(b) Now we need to form the triangles and for this we need at least three point to be drawn
For this we need to take the three cases
Case 1
If there are $2$ collinear points and one random point then the number of triangles possible would be
${}^{15}{C_1} \times {}^5{C_2}$
Here it is ${}^{15}{C_1}$ as we need to choose one random point from $15$ points and ${}^5{C_2}$ because we have to choose two point from the collinear point which are five.
Hence we get number of the triangles
$
   = {}^{15}{C_1} \times {}^5{C_2} \\
   = 15 \times \dfrac{{5!}}{{3!2!}} = 15(10) = 150 \\
 $
Case 2
If there are $1$ collinear points and two random point then the number of triangles possible would be
${}^{15}{C_2} \times {}^5{C_1}$
Here it is ${}^5{C_1}$ as we need to choose one point from $5$ collinear points and ${}^{15}{C_2}$ because we have to choose two random points.
Hence we get number of the triangles
$
   = {}^{15}{C_2} \times {}^5{C_1} \\
   = \dfrac{{15!}}{{13!2!}} \times 5 = 525 \\
 $
Case 3
If we have all three points from the random then we get that the number of the triangles as
$
   = {}^{15}{C_3} \\
   = \dfrac{{15!}}{{12!3!}} = \dfrac{{15(14)(13)}}{6} = 455 \\
 $
So total number of triangle will be$ = 150 + 525 + 455 = 1130$
(c) For the quadrilateral we need minimum of the $4$ points.
Case 1
If one point is from the collinear and other $3$ are from the random then
Number of quadrilateral will be
$
   = {}^{15}{C_3} \times {}^5{C_1} \\
   = \dfrac{{15(14)(13)}}{{3(2)}} \times 5 = 2275 \\
 $
Case 2
If two point are from the collinear and other $2$ are from the random then
Number of quadrilateral will be
$
   = {}^{15}{C_2} \times {}^5{C_2} \\
   = \dfrac{{15(14)}}{{(2)}} \times \dfrac{{5(4)}}{2} = 1050 \\
 $
Case 3
If all three points are from the random then
Number of quadrilateral will be
$
   = {}^{15}{C_4} \\
   = \dfrac{{15(14)(13)(12)}}{{(4)(3)(2)}} = 1365 \\
 $

So the total number of quadrilaterals$ = 2275 + 1050 + 1365 = 4690$

Note:
If there are total of n points in the plane out of which $m$ are collinear then we can say that number of triangles formed is given by the formula \[{}^n{C_3} - {}^m{C_3}\] as we need at least three points to form a triangle.