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Question

There are $$m$$ points in a plane which are joined by straight lines in all possible ways and of these no two are coincident or parallel and no three of them are concurrent except at the points. Show that the number of points of intersection, other than the given points, of the lines are formed is/are $$\displaystyle\frac{m!}{8(m-4)!}$$


A
m!8(m4)!
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B
m+1!2(m4)!
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C
m!4(m4)!
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D
None of these
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Solution

The correct option is D $$\displaystyle\frac{m!}{8(m-4)!}$$
Let $$m$$ points be $$A_{1}, A_{2},A_{3}, ... A_{m}$$
We consider four points $$A_{1}, A_{2},A_{3}, A_{4}$$
If these points are joined in all possible cases we have 3 points of intersection $$H_{1}, H_{2} and H_{3}$$
So the required points of intersection $$= 3.^mC_4$$
$$= \displaystyle 3.\frac{m!}{4!(m-4)!}
=\frac{m!}{8( m-4)!}$$
Hence, option 'A' is correct.

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