Question: There are $ n $ straight lines in a plane, no two of which are parallel and no three pass through the same point. Their points of intersection are joined. Then the number of fresh lines thus obtained is
Options:
A) $ \frac{n(n-1)(n-2)}{8} $
B) $ \frac{n(n-1)(n-2)(n-3)}{6} $
C) $ \frac{n(n-1)(n-2)(n-3)}{8} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Since no two lines are parallel and no three are concurrent, therefore $ n $ straight lines intersect at $ ^{n}C_2=N $ (say) points. Since two points are required to determine a straight line, therefore the total number of lines obtained by joining $ N $ points $ ^{N}C_2 $ . But in this each old line has been counted $ ^{n-1}C_2 $ times, since on each old line there will be $ n-1 $ points of intersection made by the remaining $ (n-1) $ lines. Hence the required number of fresh lines is $ ^{N}C_2-n\ .{{\ }^{n-1}}C_2=\frac{N(N-1)}{2}-\frac{n(n-1)(n-2)}{2} $ $ =\frac{^{n}C_2{{(}^{n}}C_2-1)}{2}-\frac{n(n-1)(n-2)}{2}=\frac{n(n-1)(n-2)(n-3)}{8} $ .
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