Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Determine the largest number of triangles that may be formed by selecting vertices from a collection of 27 points, 15 of which are on the same straight line.<s

Determine the largest number of triangles that may be formed by selecting vertices from a collection of 27 points, 15 of which are on the same straight line.

Option: 1

2375
 


Option: 2

1375 


Option: 3

2470 

 


Option: 4

1270


Answers (1)

best_answer

Three points are required to make a triangle, hence 27 points result in ^{27}\mathrm{C}_{3} triangles.

The total number of triangles formed from 27 points =^{27}\mathrm{C}_{3}

Thus,

\mathrm{\begin{aligned} &{ }^{27} C_3=\frac{27 !}{3 ! 24 !}\\ &{ }^{27} C_3=\frac{27 \times 26 \times 25}{3 \times 2}\\ &{ }^{27} C_3=2925 \end{aligned}}

However, in a statement that was made very clear, 15  points were in the same line and formed a triangle in ^{15}\mathrm{C}_{3} ways.

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 }

It is known that the 15 points are on the same line.

So, the 15 points out of 27 are given by,

\mathrm{ { }^{27} C_3-{ }^{15} C_3=2925-455 }

\mathrm{{ }^{27} C_3-{ }^{15} C_3=2470}

Therefore, the number of triangles formed is 2470 .

 

Posted by

Gaurav

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: NIT Rourkela cut-offs for CSE, chemical engineering drop; last years' closing ranks
anam-khan

JEE Main 2025 Live: NTA exam city slip at jeemain.nta.nic.in soon; session 1 schedule
anam-khan

JEE Main 2025 schedule out for session 1; paper 1 from January 22

Latest Question