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Question: If 7 points out of 12 are in the same straight line. Then the number of triangles formed is: A.19 ...

If 7 points out of 12 are in the same straight line. Then the number of triangles formed is:
A.19
B.158
C.185
D.201

Explanation

Solution

This question comes under the concept of combinations because, the question is about selecting the items from the samples and not arranging which is permutations, A triangle can be formed by 3 points and it can be done in 12C3^{12}{C_3}ways and out of 12 points 7 are in straight line, where triangle cannot be formed. So, the difference of 12C3^{12}{C_3}and the combination of triangles which cannot be formed by 7 points is the final answer.

Complete step-by-step answer:
A triangle can be formed by joining 3 points out of 12 points, these 12 points include 7 points which are in a straight line.
So, the combination in which the number of triangles that can be formed by using 12 points is 12C3^{12}{C_3}ways.
These 12 points contain 7 points which are in a straight line, where a triangle cannot be formed in 7C3^7{C_3}ways.
So, the number of triangles formed are,

12C37C3 12!3!(123)!7!3!(73)! 12!3!9!7!3!4! 12×11×10×9!3!×9!7×6×5×4!3!×4! 12×11×103×2×17×6×53×2×1 22035 185  { \Rightarrow ^{12}}{C_3}{ - ^7}{C_3} \\\ \Rightarrow \dfrac{{12!}}{{3!\left( {12 - 3} \right)!}} - \dfrac{{7!}}{{3!\left( {7 - 3} \right)!}} \\\ \Rightarrow \dfrac{{12!}}{{3!9!}} - \dfrac{{7!}}{{3!4!}} \\\ \Rightarrow \dfrac{{12 \times 11 \times 10 \times 9!}}{{3! \times 9!}} - \dfrac{{7 \times 6 \times 5 \times 4!}}{{3! \times 4!}} \\\ \Rightarrow \dfrac{{12 \times 11 \times 10}}{{3 \times 2 \times 1}} - \dfrac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \\\ \Rightarrow 220 - 35 \\\ \Rightarrow 185 \\\

So, the number of triangles formed are 185 triangles from 12 points, where 7 points are in a straight line.
So, option C is correct.

Note: When question is about formation, selection go for combinations, when it is for arrangement go for permutations. Understand the concept before attempting permutation and combination questions.