There are 15 points in a plane of which 10 are collinear how many triangles can be formed

Answer

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Hint:
A triangle is formed by joining 3 points, this is a concept of combination. A triangle can be formed when only 2 points are collinear, if more than 2 points are collinear then the triangle cannot be formed.

Complete step by step solution:
Before going into the solution, let us be clear about the concept
Permutation: Arranging the numbers in order is called permutation, the formula of permutation is \[^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Where n= Total number of items in the sample, r= number of items to be selected from the sample.
Combination: Selecting the items from the sample is called combination, the formula of combination is \[^n{C_r} = {\dfrac{{n!}}{{r!\left( {n - r} \right)!}}_r}\]
Where n= Total number of items in the sample, r= number of items to be selected from the sample.
1) No three of which are collinear
Given that no three of which are collinear, which means no three points are not in a straight line which means out of 10 points, the number of triangles can be formed in $^{10}{C_3}$ ways.
Number of triangles that can be formed when no three of which are collinear are,
\[
  { \Rightarrow ^{10}}{C_3} \\
   \Rightarrow \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}} \\
   \Rightarrow \dfrac{{10!}}{{3!7!}} \\
   \Rightarrow \dfrac{{10 \times 9 \times 8 \times 7!}}{{3!7!}} \\
   \Rightarrow \dfrac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}} \\
   \Rightarrow 120 \\
 \]
The number of triangles that can be formed is in 120 ways.
2) Four points are collinear.
Given that four points are collinear, which means four points are in the straight line, whereas a triangle needs three points to form a triangle.Out of 10 points, 4 points cannot form a triangle. So, the number of triangles formed when four points are collinear are in $^{10}{C_3}{ - ^4}{C_3}$ways.
Number of triangles that can be formed when four points are collinear are
\[
  { \Rightarrow ^{10}}{C_3}{ - ^4}{C_3} \\
   \Rightarrow \dfrac{{10!}}{{3!\left( {10 - 3} \right)!}} - \dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} \\
   \Rightarrow \dfrac{{10 \times 9 \times 8 \times 7!}}{{3!7!}} - \dfrac{{4 \times 3 \times 2 \times 1!}}{{3!1!}} \\
   \Rightarrow \dfrac{{10 \times 9 \times 8}}{{3 \times 2 \times 1}} - \dfrac{{4 \times 3 \times 2}}{{3 \times 2 \times 1}} \\
   \Rightarrow 120 - 4 \\
   \Rightarrow 116 \\
 \]$$

The number of triangles that can be formed when Four points are collinear are 116.

Note:
When points are collinear, a triangle cannot be formed, unless if only two points are collinear. When all points are in collinear no triangle can be formed. Because it won’t be having any area. That’s why we use it to show the area of a triangle is zero to prove points are collinear.

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Concept:

Combination formula:

The formula of a combination of r objects out of n objects is given as follows:

\(\rm ^nC_r = \dfrac{n!}{r!(n-r)!}\)

Calculation:

There are 15 points in a plane the triangle formed by joining the three-point

Therefore,

Number of triangle = 15C3

= \(\dfrac{15!}{3!(15-3)!}\)

= \(\dfrac{15\times14\times13}{3\times2}\)

= 455

Last updated on Sep 22, 2022

The exam dates for the HTET 2022 have been postponed. Due to the General Elections, the exam dates for the HTET have been revised. The exam will be conducted on the 3rd and 4th of December 2022 instead of the 12th and 13th of November 2022. The exam is conducted by the Board of School Education, Haryana to shortlist eligible candidates for PGT and TGT posts in Government schools across Haryana. The exam is conducted for 150 marks. The HTET Exam Pattern for Level I, Level II, and Level III exams is different. There will be no negative marking in the exam.

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Solution : The number of lines that can be formed from n points in which m points are collinear is `.^(n)C_(2)-.^(m)C_(2)+1`.

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