Question

Number of points of intersection of n straight lines if n satisfies ${}^{n+5}{{P}_{n+1}}=\dfrac{11\left( n-1 \right)}{2}\times {}^{n+3}{{P}_{n}}$ is:
(a). 15
(b). 28
(c). 21
(d). 10

Answer 1

Hint: First look at the definition of P and then substitute the value of P. Now you get an equation with n and factorial terms. Now just cancel the common factorials on both sides. By this you get a quadratic in n. with variables on both sides. Now try to make the right hand side as zero. And then find the value of n we know 2 lines intersect at one point. So, from n lines number of 2 line pairs will be number of intersection points ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$

Complete step-by-step solution -
Permutations: In mathematics, permutations of a set is an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word permutation itself refers to out or process of changing the linear order of an ordered set. Its formula is given by : ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.
Factorial:- In mathematics, factorial is an operation, denoted by “$\left( ! \right)$ “. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as: $n!=n\times \left( n-1 \right)\times \left( n-2 \right)...................\times 1$
For example: $5!=5\times 4\times 3\times 2\times 1=120.$
Note that we assume $0!=1$ . It is standard value. It has a wide range of applications in combinatorics.
Given condition in the question, in terms of n, is given by ${}^{n+5}{{P}_{n+1}}=\dfrac{11\left( n-1 \right)}{2}\times {}^{n+3}{{P}_{n}}$ .
By substituting the formula we can write the equation as: $\dfrac{\left( n+5 \right)!}{\left( n+5-n-1 \right)!}=\dfrac{11\left( n-1 \right)}{2}\times \dfrac{\left( n+3 \right)!}{\left( n+3-n \right)!}$
By simplifying the above equation, we can write it as: $\dfrac{\left( n+5 \right)!}{4!}=\dfrac{11\left( n-1 \right)}{2}\times \dfrac{\left( n+3 \right)!}{3!}$ .
By looking at terms on the right hand side, we can write it as: (use definition of factorial). $\dfrac{\left( n+5 \right)\left( n+4 \right)}{4}\dfrac{\left( n+3 \right)!}{3!}=\dfrac{11\left( n-1 \right)}{2}\times \dfrac{\left( n+3 \right)!}{3!}$ .
By simplifying the above equation, we get it as: ${{n}^{2}}+5n+4n+20=22\left( n-1 \right)$.
By subtracting $22\left( n-1 \right)$ on both sides, we get it as: ${{n}^{2}}-13n+42=0$ .
We can write the equation as a form of: ${{n}^{2}}-6n-7n+42=0$ .
By factoring the above equation, we get: $\left( n-6 \right)\left( n-7 \right)=0$ .
So, the roots of n are 6 and 7.
We know 2 straight lines form one intersection point.
So, if there are n straight lines. Then the number of ways to select 2 from these n will give a number of intersections.
So, the number of intersections $={}^{n}{{C}_{2}}$ .
So, if $n=7$ , we get intersections as ${}^{7}{{C}_{2}}=\dfrac{7!}{2!5!}=21$.
So, if $n=6$ , we get an intersection as ${}^{6}{{C}_{2}}=\dfrac{6!}{2!4!}=15$ .
So, in a given condition the number of intersections are 21, 15.
Therefore options (a) (c) are correct.

Note: Here these are 2 possibilities. Students generally solve for one and forget the other, common terms are canceled here because we know factorials are never zero. Don’t solve the factorial completely. Solve only till a point where you get the common term for the right hand side of the equation to cancel with it.