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Q.
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
Permutations and Combinations
Solution:
${ }^{n+5} P_{n+1}=\frac{11(n-1)}{2} \times{ }^{n+3} P_n$
or ${ }^{n+5} P_{n+1}=\frac{(n+5)!}{4!}=\frac{11(n-1)}{2} \frac{(n+3)!}{3!}$
or $(n+5)(n+4)=22(n-1)$
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 $21$.