Question

There are three coplanar parallel lines. If any p points are taken on each line, the maximum number of triangles with vertices at these points is

Answer 1

Count the number of ways of choosing any 3 points from all the points, i.e., 3p. Count the number of ways of choosing 3 collinear points from all the lines. Subtract the number of ways of choosing 3 collinear points from the number of ways of choosing any 3 points from all the points.

Complete step-by-step answer:
We have three coplanar parallel lines such that each line has p points. We have to calculate the maximum number of triangles which can be formed with vertices at these points.
To calculate the maximum number of triangles which can be formed, we have to calculate the number of ways of choosing 3 vertices such that they form a triangle.
We will firstly count the number of ways of choosing 3 points from all the points.
As there are p points on each line and there are 3 lines, total number of points $=p+p+p=3p$. We have to choose any 3 points from these 3p points.
We know that we can choose ‘r’ objects from a set of ‘n’ objects in ${}^{n}{{C}_{r}}$ ways. Thus, we can choose 3 points from 3p points in ${}^{3p}{{C}_{3}}$ ways.
But we observe that we can’t form a triangle if all the three points are collinear. So, we have to remove the number of 3 collinear points from each line.
As there are p points on each line, there are a number of ways to choose 3 points from these p points $={}^{p}{{C}_{3}}$.
To calculate the number of ways to choose 3 collinear points from all the lines $=3\left( {}^{p}{{C}_{3}} \right)$. We will subtract this number from the total number of ways to choose 3 points to count the number of ways of choosing 3 vertices such that they form a triangle.
Thus, number of ways to choose 3 vertices that form a triangle $={}^{3p}{{C}_{3}}-3\left( {}^{p}{{C}_{3}} \right)$.
Simplifying the above expression, maximum number of possible triangles $=\dfrac{\left( 3p \right)!}{3!\left( 3p-3 \right)!}-3\left( \dfrac{p!}{3!\left( p-3 \right)!} \right)$.
Thus, the maximum number of possible triangles $=\dfrac{3p\left( 3p-1 \right)\left( 3p-2 \right)}{6}-\dfrac{3p\left( p-1 \right)\left( p-2 \right)}{6}=\dfrac{3p}{6}\left[ \left( 3p-1 \right)\left( 3p-2 \right)-\left( p-1 \right)\left( p-2 \right) \right]$.
Further simplifying the above expression, the maximum number of possible triangles $=\dfrac{p}{22}\left[ \left( 9{{p}^{2}}-6p-3p+2 \right)-\left( {{p}^{2}}-2p-p+2 \right) \right]=\dfrac{p}{2}\left( 9{{p}^{2}}-{{p}^{2}}-9p+3p+2-2 \right)$.
So, the number of possible triangles $=\dfrac{p}{2}\left( 8{{p}^{2}}-6p \right)=p\left( 4{{p}^{2}}-3p \right)={{p}^{2}}\left( 4p-3 \right)$.
Hence, the maximum number of possible triangles with vertices on these lines is ${{p}^{2}}\left( 4p-3 \right)$.

Note: It’s necessary to remove the number of ways to choose 3 collinear points from the total number of ways to choose 3 points as joining any 3 collinear points will give us a straight line. Also, one must know that ${}^{n}{{C}_{r}}$ can be simplified using ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.