Question

There are n straight lines in a plane, in which no two are parallel and no three pass through the same point. Their points of intersection are joined. Show that the number of fresh lines thus introduced is $\dfrac{1}{8}n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)$.

Answer 1

Hint: This is a question based on permutations and combinations. Here you need to focus on the point that there is only one unique line passing through 2 given points. Also, you need to take the point into consideration that the points formed by the intersection of lines will definitely consist of collinear points as well.

Complete step-by-step answer:

Let us start the solution by finding the total number of points of intersection we will have when n straight lines intersect. So, basically to get the number of points of intersection we need to select 2 lines at a time, as two lines will have a unique point of intersection between them. Therefore, the total number of points of intersection is equal to $^{n}{{C}_{2}}=\dfrac{n(n-1)}{2}$ .

Now if you try to think a bit on the situation, you will notice that if we pick a line out of the n old lines, a total of n-1 points of intersection lie on it and if we join any two points out of this n-1 point we will get the old line only and this will happen for all n lines. Therefore, we can say that the number of new lines formed is:

$^{\dfrac{n\left( n-1 \right)}{2}}{{C}_{2}}-{{n}^{n-1}}{{C}_{2}}$   

Here in first term we selected 2 point of intersection out of the total number of points of intersection($^{n}{{C}_{2}}=\dfrac{n(n-1)}{2}$) and the second term is subtracted because  if we join any two points out of this n-1 point we will get the old line only and this will happen for all n lines.

The first term in the expression is the number of possible ways of selecting 2 points out of the $^{n}{{C}_{2}}=\dfrac{n(n-1)}{2}$ point of intersections. While the second term is the older lines we get due to each collection of n-1 points of intersections as explained above. 

So, if we solve the expression, we get

$^{\dfrac{n\left( n-1 \right)}{2}}{{C}_{2}}-{{n}^{n-1}}{{C}_{2}}=\dfrac{_{\left( \dfrac{n(n-1)}{2} \right)\left( \dfrac{n(n-1)}{2}-1 \right)}}{2}-n\times \dfrac{(n-1)(n-2)}{2}$

${{\Rightarrow }^{\dfrac{n\left( n-1 \right)}{2}}}{{C}_{2}}-{{n}^{n-1}}{{C}_{2}}=\dfrac{n(n-1)({{n}^{2}}-n-2)}{8}-n\times \dfrac{(n-1)(n-2)}{2}$

${{\Rightarrow }^{\dfrac{n\left( n-1 \right)}{2}}}{{C}_{2}}-{{n}^{n-1}}{{C}_{2}}=n(n-1)\left( \dfrac{{{n}^{2}}-n-2-4(n-2)}{8} \right)$

${{\Rightarrow }^{\dfrac{n\left( n-1 \right)}{2}}}{{C}_{2}}-{{n}^{n-1}}{{C}_{2}}=n(n-1)\left( \dfrac{{{n}^{2}}-5n+6}{8} \right)$

${{\Rightarrow }^{\dfrac{n\left( n-1 \right)}{2}}}{{C}_{2}}-{{n}^{n-1}}{{C}_{2}}=n(n-1)\left( \dfrac{{{n}^{2}}-2n-3n+6}{8} \right)$

${{\Rightarrow }^{\dfrac{n\left( n-1 \right)}{2}}}{{C}_{2}}-{{n}^{n-1}}{{C}_{2}}=n(n-1)\dfrac{(n-2)(n-3)}{8}$

${{\Rightarrow }^{\dfrac{n\left( n-1 \right)}{2}}}{{C}_{2}}-{{n}^{n-1}}{{C}_{2}}=\dfrac{1}{8}n(n-1)(n-2)(n-3)$

So, looking at the result, we can say that we have proved the point asked in the question.

Note: In questions related to permutations and combinations, students generally face problems in deciding whether the question is based on selection or arrangement. Also, you might face problems in analysing whether to make cases or not. Also, be careful about the conditions mentioned in the question, as terms like non-parallel, non-collinear etc. are very crucial in such questions.