Question

Number of points of intersection of $n$ straight lines of $n$ satisfies ${}^{n+5}{P}_{n+1}=\frac{11(n-1)}{2}{\times }^{n+3}{P}_n$ is

• A. $15$
• B. $28$
• C. $21$
• D. $10$

Answer 1

Answer: (A), (C)

$15$
$21$

${}^{n+5}{P}_{n+1}=\frac{11(n-1)}{2}{\times }^{n+3}{P}_n\Rightarrow \frac{(n+5)!}{4!}=\frac{11(n-1)}{2}\times \frac{(n+3)!}{3!}$$(n+5)(n+4)=22(n-1)\Rightarrow n=7or6$
Thus, we have number of intersections: For each pair of straight lines, we have 1 intersection.
Hence, no. of intersection = ${}^6{C}_2=15or{}^7{C}_2=21$
Hence, (a) and (c) are correct.