Permutations and CombinationsMCQMTP Nov 20Question 1729 of 251
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The number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the same straight line, is:

Options

A185
B175
C115
D105
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Correct Answer

Option a185

All Options:

  • A185
  • B175
  • C115
  • D105

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Detailed Solution & Explanation

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

If all 12\displaystyle 12 points were in general position (no three collinear), the number of ways to choose 3\displaystyle 3 points is:
12C3=12×11×103×2×1=220^{12}C_{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220
Since 7\displaystyle 7 of these points lie on the same straight line, any selection of 3\displaystyle 3 points from these 7\displaystyle 7 collinear points will fail to form a triangle (they will form a straight line instead). The number of ways to choose 3\displaystyle 3 points from these 7\displaystyle 7 collinear points is:
7C3=7×6×53×2×1=35^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
The total number of triangles that can be formed is:
Number of Triangles=12C37C3\text{Number of Triangles} = ^{12}C_{3} - ^{7}C_{3}
Number of Triangles=22035=185\text{Number of Triangles} = 220 - 35 = 185

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.

Key Concepts to Understand

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