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(B) Given $n = 20$ straight lines in a plane,where no two are parallel and no three are concurrent. The number of intersection points is given by $\bi

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$14535$

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(B) Given $n = 20$ straight lines in a plane,where no two are parallel and no three are concurrent. The number of intersection points is given by $\binom{n}{2} = \binom{20}{2} = \frac{20 	imes 19}{2} = 190$. Let $I = 190$ be the number of intersection points. Each pair of these intersection points forms a line segment. The total number of line segments formed by joining these $I$ points is $\binom{I}{2}$. However,we must exclude the segments that lie on the original $20$ lines. For each line,there are $n-1 = 19$ intersection points. The number of segments on one line is $\binom{19}{2}$. Since there are $20$ lines,the number of segments to exclude is $20 	imes \binom{19}{2} = 20 	imes \frac{19 	imes 18}{2} = 20 	imes 171 = 3420$. The number of new line segments is $\binom{190}{2} - 3420 = \frac{190 	imes 189}{2} - 3420 = 17955 - 3420 = 14535$.

There are $20$ straight lines in a plane such that no two of them are parallel and no three of them are concurrent. If their points of intersection are joined,then the number of new line segments formed is

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