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Find the maximum number of triangles that may be created by selecting the vertices from a group of 15 points, ten of which are on the same straight line.

 

Option: 1

335
 


Option: 2

225


Option: 3

445

 


Option: 4

525


Answers (1)

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Three points are required to make a triangle, hence 15 points result in \mathrm{^{15}C_{3}} triangles.

The total number of triangles formed from 15 points = \mathrm{^{15}C_{3}}

Thus,

\mathrm{ { }^{15} C_3=\frac{15 !}{3 ! 12 !} }

\mathrm{ { }^{15} C_3=\frac{15 \times 14 \times 13}{3 \times 2} }

\mathrm{ { }^{15} C_3=455}

However, in a statement that was made very clear, 10 points were in the same line and formed a triangle in \mathrm{ { }^{10} C_3} ways.

Thus,

\mathrm{\begin{aligned} &{ }^{10} C_3=\frac{10 !}{3 ! 7 !}\\ &{ }^{10} C_3=\frac{10 \times 9 \times 8}{3 \times 2}\\ &{ }^{10} C_3=120 \end{aligned}}

It is known that the 10 points are on the same line.

So, the 10 points out of 15 are given by, 

\mathrm{ { }^{15} C_3-{ }^{10} C_3=455-120 }

\mathrm{{ }^{15} C_3-{ }^{10} C_3=335}

Therefore, the number of triangles formed is 335.

 

 

 

Posted by

Ritika Jonwal

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