Question

Maximum number of points of intersection of $n$ straight lines if $n$ 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)

$21$

Given
${}^{n+5}{P}_{n+1}=\frac{11(n-1)}{2}\times {}^{n+3}{P}_n$

$\Rightarrow {}^{n+5}{P}_{n+1}\Rightarrow \frac{(n+5)!}{4!}=\frac{11(n-1)}{2}\frac{(n+3)!}{3!}$

$\Rightarrow (n+5)(n+4)=22(n-1)$

After solving, we get $n=6$ or $n=7$.
The number of points of intersection of lines is ${}^6{C}_2$ or ${}^7{C}_2\equiv 15$ or $21$