Permutations and CombinationsMCQMTP Nov 20Question 1729 of 249
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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:

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option a — 185

All Options:

  • A185
  • B175
  • C115
  • D105

Detailed Solution & Explanation

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

If all $12$ points were in general position (no three collinear), the number of ways to choose $3pointsis:<br><spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><msup><mrow></mrow><mn>12</mn></msup><msub><mi>C</mi><mn>3</mn></msub><mo>=</mo><mfrac><mrow><mn>12</mn><mo>×</mo><mn>11</mn><mo>×</mo><mn>10</mn></mrow><mrow><mn>3</mn><mo>×</mo><mn>2</mn><mo>×</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mn>220</mn></mrow><annotationencoding="application/x−tex">12C3=12×11×103×2×1=220</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:1.0141em;vertical−align:−0.15em;"></span><spanclass="mord"><span></span><spanclass="msupsub"><spanclass="vlist−t"><spanclass="vlist−r"><spanclass="vlist"style="height:0.8641em;"><spanstyle="top:−3.113em;margin−right:0.05em;"><spanclass="pstrut"style="height:2.7em;"></span><spanclass="sizingreset−size6size3mtight"><spanclass="mordmtight"><spanclass="mordmtight">12</span></span></span></span></span></span></span></span></span><spanclass="mord"><spanclass="mordmathnormal"style="margin−right:0.07153em;">C</span><spanclass="msupsub"><spanclass="vlist−tvlist−t2"><spanclass="vlist−r"><spanclass="vlist"style="height:0.3011em;"><spanstyle="top:−2.55em;margin−left:−0.0715em;margin−right:0.05em;"><spanclass="pstrut"style="height:2.7em;"></span><spanclass="sizingreset−size6size3mtight"><spanclass="mordmtight"><spanclass="mordmtight">3</span></span></span></span></span><spanclass="vlist−s">​</span></span><spanclass="vlist−r"><spanclass="vlist"style="height:0.15em;"><span></span></span></span></span></span></span><spanclass="mspace"style="margin−right:0.2778em;"></span><spanclass="mrel">=</span><spanclass="mspace"style="margin−right:0.2778em;"></span></span><spanclass="base"><spanclass="strut"style="height:2.0908em;vertical−align:−0.7693em;"></span><spanclass="mord"><spanclass="mopennulldelimiter"></span><spanclass="mfrac"><spanclass="vlist−tvlist−t2"><spanclass="vlist−r"><spanclass="vlist"style="height:1.3214em;"><spanstyle="top:−2.314em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">3</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mbin">×</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mord">2</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mbin">×</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mord">1</span></span></span><spanstyle="top:−3.23em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="frac−line"style="border−bottom−width:0.04em;"></span></span><spanstyle="top:−3.677em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">12</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mbin">×</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mord">11</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mbin">×</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mord">10</span></span></span></span><spanclass="vlist−s">​</span></span><spanclass="vlist−r"><spanclass="vlist"style="height:0.7693em;"><span></span></span></span></span></span><spanclass="mclosenulldelimiter"></span></span><spanclass="mspace"style="margin−right:0.2778em;"></span><spanclass="mrel">=</span><spanclass="mspace"style="margin−right:0.2778em;"></span></span><spanclass="base"><spanclass="strut"style="height:0.6444em;"></span><spanclass="mord">220</span></span></span></span></span><br>Since\displaystyle points is:<br><span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><msup><mrow></mrow><mn>12</mn></msup><msub><mi>C</mi><mn>3</mn></msub><mo>=</mo><mfrac><mrow><mn>12</mn><mo>×</mo><mn>11</mn><mo>×</mo><mn>10</mn></mrow><mrow><mn>3</mn><mo>×</mo><mn>2</mn><mo>×</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mn>220</mn></mrow><annotation encoding="application/x-tex">^{12}C_{3} = 12×11×103×2×1=220\displaystyle \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:1.0141em;vertical-align:-0.15em;"></span><span class="mord"><span></span><span class="msupsub"><span class="vlist-t"><span class="vlist-r"><span class="vlist" style="height:0.8641em;"><span style="top:-3.113em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">12</span></span></span></span></span></span></span></span></span><span class="mord"><span class="mord mathnormal" style="margin-right:0.07153em;">C</span><span class="msupsub"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:0.3011em;"><span style="top:-2.55em;margin-left:-0.0715em;margin-right:0.05em;"><span class="pstrut" style="height:2.7em;"></span><span class="sizing reset-size6 size3 mtight"><span class="mord mtight"><span class="mord mtight">3</span></span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.15em;"><span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:2.0908em;vertical-align:-0.7693em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.3214em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">3</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">2</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">1</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">12</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">11</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">10</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.7693em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">220</span></span></span></span></span><br>Since7$ of these points lie on the same straight line, any selection of $3$ points from these $7$ collinear points will fail to form a triangle (they will form a straight line instead). The number of ways to choose $3$ points from these $7$ collinear points is:
7C3=7×6×53×2×1=35^{7}C_{3} = 7×6×53×2×1=35\displaystyle \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=12C3−7C3Number of Triangles=12C3−7C3\displaystyle \text{Number of Triangles} = ^{12}C_{3} - ^{7}C_{3}
Number of Triangles=220−35=185Number of Triangles=220−35=185\displaystyle \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

Set

A well-defined collection of distinct objects (called elements or members). Sets are denoted by capital letters and elements are listed within curly braces. Key operations include Union (∪), Intersection (∩), and Complement.

Related Comparison Tables

Permutations vs Combinations: The Quantitative Aptitude Guide

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