Question

The straight lines \[{{L}_{1}}\], \[{{L}_{2}}\], \[{{L}_{3}}\] are parallel and lie in the same plane. A total number of m points are taken on \[{{L}_{1}}\] , n points on \[{{L}_{2}}\] and k points on \[{{L}_{3}}\]. Then maximum number of triangles formed with vertices at these points are
A. \[{}^{m+n+k}{{C}_{3}}\]
B. \[{}^{m+n+k}{{C}_{3}}-({}^{m}{{C}_{3}}+{}^{n}{{C}_{3}}+{}^{k}{{C}_{3}})\]
C. \[({}^{m}{{C}_{3}}+{}^{n}{{C}_{3}}+{}^{k}{{C}_{3}})\]
D. \[{}^{m}{{C}_{2}}.{}^{n}{{C}_{2}}+{}^{n}{{C}_{1}}.{}^{k}{{C}_{2}}+{}^{n}{{C}_{2}}.{}^{m}{{C}_{1}}\]

Answer 1

Hint: To form a triangle we will require 3 points in such a way that these points are not collinear, it means all the 3 points should not be chosen from a single line. In this problem we will select three points from the given points on lines \[{{L}_{1}}\], \[{{L}_{2}}\] and \[{{L}_{3}}\] and make sure the three points are not collinear.

Complete step-by-step answer:
Total number of points given here on lines \[{{L}_{1}}\], \[{{L}_{2}}\] and \[{{L}_{3}}\] \[=m+n+k\].
Now to get the total number of possible triangles which can be formed the \[m+n+k\] points, we will use the concept of combination. Then, we can write it as \[{}^{m+n+k}{{C}_{3}}\].
We have to note here that this total number of possible triangles \[\left( {}^{m+n+k}{{C}_{3}} \right)\] also includes triangles which are not possible to form, for example it also includes the 3 points taken from the same line which doesn’t form a triangle. Like 3 points are selected from m points on \[{{L}_{1}}\] will not form a triangle as all the three points are collinear. We will subtract these cases from the total number of possible triangles which can be formed.

Cases which produce no triangles are

Case I: Taking 3 points from m points on line \[{{L}_{1}}\]
We will get the number of such cases as \[={}^{m}{{C}_{3}}\]

Case II: Taking 3 points from n points on line \[{{L}_{2}}\]
We will get the number of such cases as \[={}^{n}{{C}_{3}}\]

Case III: Taking 3 points from n points on line \[{{L}_{3}}\]
We will get the number of such cases as \[={}^{k}{{C}_{3}}\]
Now, we can add the results of these three cases to get the total number of cases which produce no triangles as \[{}^{m}{{C}_{3}}+{}^{n}{{C}_{3}}+{}^{k}{{C}_{3}}\].
Therefore, we get the maximum number of triangles formed \[={}^{m+n+k}{{C}_{3}}-({}^{m}{{C}_{3}}+{}^{n}{{C}_{3}}+{}^{k}{{C}_{3}})\]
Therefore, option B is the correct answer.

Note: In this type of problem just keep in mind that to form a triangle, 3 points should be chosen in a way that those points are not collinear. If students ignore this particular point, then they will end up choosing option A. \[{}^{m+n+k}{{C}_{3}}\] as the correct answer. We use combinations when we have to choose ‘r’ number of objects from ‘n’ number of objects. It is represented as $ {}^{n}{{C}_{r}} $ and the formula is given by $ {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ .