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A plane has nine points in it. Four of these nine points are in a straight line, and except for these four points, no other 2 points are in the same straight line. Find how many quadrilaterals can be created by connecting these 9 points.

 

Option: 1

115
 


Option: 2

120


Option: 3

125

 


Option: 4

135


Answers (1)

best_answer

Given that,

There are 9 points in a plane.

The quadrilateral needs 4 points.

Case 1:

No. of points selected out of the remaining 5 points = 4

No. of points selected out of 4 collinear points = 0

Thus,

The number of quadrilaterals formed is,

\mathrm{{ }^4 C_0 \times{ }^5 C_4=5}

Case 2:

No. of points selected out of the remaining 5 points = 3

No. of points selected out of 4 collinear points = 1

Thus, 

The number of quadrilaterals formed is,

\mathrm{\begin{aligned} & { }^4 C_1 \times{ }^5 C_3=4 \times \frac{5 !}{3 ! 2 !} \\ & { }^4 C_1 \times{ }^5 C_3=40 \end{aligned}}

Case 3:

No. of points selected out of the remaining 5 points = 2

No. of points selected out of 4 collinear points = 2

Thus, 

The number of quadrilaterals formed is,

\mathrm{\begin{aligned} &{ }^4 C_2 \times{ }^5 C_2=\frac{4 !}{2 ! 2 !} \times \frac{5 !}{2 ! 3 !}\\ &{ }^4 C_2 \times{ }^5 C_2=60 \end{aligned}}

Case 4:

No. of points selected out of the remaining 5 points = 1

No. of points selected out of 4 collinear points = 3

Thus, 

The number of quadrilaterals formed is,

\mathrm{\begin{aligned} & { }^4 C_3 \times{ }^5 C_1=5 \times 4 \\ & { }^4 C_3 \times{ }^5 C_1=20 \end{aligned}}

Case 5:

No. of points selected out of the remaining 5 points = 0

No. of points selected out of 4 collinear points = 4

Thus, 

The number of quadrilaterals formed is,      

\mathrm{{ }^4 C_4 \times{ }^5 C_0=0}

Therefore, the total number of quadrilaterals formed is 5 + 40 + 60 +20 = 125.

Posted by

Pankaj

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