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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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Out of 18 points in a plane,no three are in the same line except five points which are collinear.Find the number of lines that can be formed joining the points

$\begin{array}{1 1}(A)\;140\\(B)\;141\\(C)\;143\\(D)\;144\end{array} $

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  • $n(C,r)=\large\frac{n!}{r!(n-r)!}$
Given :
Total points =18
Collinear points =5
No of lines formed from 18 points =$18C_2$
(Since we require two points to form one line.)
Five points are collinear =$5C_2$
$\Rightarrow $Required no of straight line =$18C_2-5C_2+1$
$\Rightarrow \large\frac{18\times 17\times 16!}{2\times 1\times 16!}$
$\Rightarrow 162$
$\Rightarrow \large\frac{5\times 4\times 3!}{2\times 3!}$
$\Rightarrow 10$
$\Rightarrow 143+1$
$\Rightarrow 144$
Hence (D) is the correct answer.
answered May 15, 2014 by sreemathi.v

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