Permutations and CombinationsMCQMTP Nov 18Question 1727 of 251
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There are 12 points in a plane which are collinear with 4 points, number of triangular that can be formed with the vertices as there points are

Options

A216
B220
C110
D108
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Correct Answer

✅ Option a — 216

All Options:

  • A216
  • B220
  • C110
  • D108

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Detailed Solution & Explanation

To form a triangle, we need to choose 3\displaystyle 3 non-collinear points.

If all 12\displaystyle 12 points in the plane were non-collinear, the total number of triangles that could be formed by choosing 3\displaystyle 3 points out of 12\displaystyle 12 would be:
Total combinations of 3 points=12C3\text{Total combinations of 3 points} = ^{12}C_{3}
12C3=12!3!(12−3)!=12×11×103×2×1=220^{12}C_{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220
However, we are given that 4\displaystyle 4 of these points are collinear (they lie on the same straight line). Any 3\displaystyle 3 points chosen from these 4\displaystyle 4 collinear points will lie on a straight line and therefore **cannot** form a triangle. The number of ways to choose 3\displaystyle 3 points from the 4\displaystyle 4 collinear points is:
Collinear combinations of 3 points=4C3=4\text{Collinear combinations of 3 points} = ^{4}C_{3} = 4
Subtracting the collinear combinations from the total combinations gives the number of valid triangles:
Number of Triangles=12C3−4C3\text{Number of Triangles} = ^{12}C_{3} - ^{4}C_{3}
Number of Triangles=220−4=216\text{Number of Triangles} = 220 - 4 = 216

Hence, **Option A** is the correct answer.

About This Chapter: Permutations and Combinations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Factorials, Permutations, Combinations

This chapter deals with the fundamental principles of counting. It covers factorials, circular permutations, restricted permutations, combinations, and the differences between selecting items versus arranging them.

View Official ICAI Syllabus

Exam Strategy Tip

The most common mistake is confusing 'P' (Arrangement) with 'C' (Selection). If order matters (like opening a lock), use P. If order doesn't matter (like choosing a team), use C.

Related Comparison Tables

Permutations vs Combinations: The Quantitative Aptitude Guide

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