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There are n points in a plane of which no three are in a straight line except ' m ' which are all in a straight line. Then the number of different quadrilaterals, that can be formed with the given points as vertices, is

Option: 1

\begin{aligned} & { }^n C_4-{ }^m C_3{ }^{n-m+1} C_1-{ }^m C_4 \\ \end{aligned}


Option: 2

{ }^n C_4-{ }^m C_3{ }^{n-m} C_1+{ }^m C_4 \\


Option: 3

{ }^n C_4-{ }^m C_3{ }^{n-m} \cdot{ }^m C_1-{ }^m C_4 \\

 


Option: 4

{ }^n C_4+{ }^n C_3 \cdot{ }^m C_1


Answers (1)

best_answer

Number of quadrilaterals = total number of quadrilaterals
                                        - choosing 3 points from m and 1 from remaining points
                                        - choosing 4 points from m points
                                            ={ }^n C_4-{ }^m C_3\left({ }^{n-m} C_1\right)-{ }^m C_4

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Divya Prakash Singh

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