$\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 =$18C_2-5C_2+1$

$18C_2=\large\frac{18!}{2!16!}$

$\Rightarrow \large\frac{18\times 17\times 16!}{2\times 1\times 16!}$

$\Rightarrow 162$

$5C_2=\large\frac{5!}{2!13!}$

$\Rightarrow \large\frac{5\times 4\times 3!}{2\times 3!}$

$\Rightarrow 10$

$18C_2-5C_2+1=153-10+1$

$\Rightarrow 143+1$

$\Rightarrow 144$

Hence (D) is the correct answer.

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