Permutations and CombinationsMCQMTP May 18Question 1706 of 249
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In how many ways 3 prizes out of 5 can be distributed amongst 3 brothers equally

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

✅ Option a — 10

All Options:

  • A10
  • B45
  • C60
  • D120

Detailed Solution & Explanation

We want to distribute 3 prizes out of 5 available prizes to 3 brothers equally (so each brother gets exactly 1 prize).
1. **Selection of prizes**:
First, we must select which 3 prizes out of the 5 will be distributed:
Ways to select prizes=5C3=5×4×33×2×1=10 ways\displaystyle \text{Ways to select prizes} = ^5C_3 = \frac{5 ×\displaystyle \times 4 ×\displaystyle \times 3}{3 ×\displaystyle \times 2 ×\displaystyle \times 1} = 10  ways\displaystyle \text{ ways}
2. **Distribution of prizes**:
Once 3 distinct prizes are selected, they can be distributed to the 3 distinct brothers in $3!\displaystyle ways:<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><mtext>Ways to distribute</mtext><mo>=</mo><mn>3</mn><mo stretchy="false">!</mo><mo>=</mo><mn>6</mn><mtext> ways</mtext></mrow><annotation encoding="application/x-tex">Ways to distribute=3\displaystyle \text{Ways to distribute} = 3! = 6  ways\displaystyle \text{ ways}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8889em;vertical-align:-0.1944em;"></span><span class="mord text"><span class="mord">Ways to distribute</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:0.6944em;"></span><span class="mord">3</span><span class="mclose">!</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:0.8389em;vertical-align:-0.1944em;"></span><span class="mord">6</span><span class="mord text"><span class="mord"> ways</span></span></span></span></span></span><br>3. **Total ways**:<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><mtext>Total</mtext><msup><mo>=</mo><mn>5</mn></msup><msub><mi>C</mi><mn>3</mn></msub><mo>×</mo><mn>3</mn><mo stretchy="false">!</mo><mo>=</mo><mn>10</mn><mo>×</mo><mn>6</mn><mo>=</mo><mn>60</mn><mtext> ways</mtext></mrow><annotation encoding="application/x-tex">Total=5C3\displaystyle \text{Total} = ^5C_3 ×\displaystyle \times 3! = 10 ×\displaystyle \times 6 = 60  ways\displaystyle \text{ ways}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8641em;"></span><span class="mord text"><span class="mord">Total</span></span><span class="mspace" style="margin-right:0.2778em;"></span><span class="mrel"><span class="mrel">=</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">5</span></span></span></span></span></span></span></span><span class="mspace" style="margin-right:0.2778em;"></span></span><span class="base"><span class="strut" style="height:0.8333em;vertical-align:-0.15em;"></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">3</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:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.6944em;"></span><span class="mord">3</span><span class="mclose">!</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:0.7278em;vertical-align:-0.0833em;"></span><span class="mord">10</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">×</span><span class="mspace" style="margin-right:0.2222em;"></span></span><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">6</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:0.8389em;vertical-align:-0.1944em;"></span><span class="mord">60</span><span class="mord text"><span class="mord"> ways</span></span></span></span></span></span><br>Mathematically, the total number of distribution ways is &#36;60 (Option C). However, the textbook answer key only considers the selection step (5C3=10\displaystyle ^5C_3 = 10) and lists Option A as correct.
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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