Permutations and CombinationsMCQMTP June 24 Series IIQuestion 1744 of 249
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The no. of ways that $12$ prizes can be divided among $5$ students so that each may give $3$ prizes is

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

✅ Option d — $16,800$

All Options:

  • A$15,400$
  • B$15,000$
  • C$14,400$
  • D$16,800$

Detailed Solution & Explanation

Let us analyze the printed question:
"The no. of ways that $12$ prizes can be divided among $5$ students so that each may give $3prizesis"<br><br>∗∗1.MathematicalImpossibilityofthePrintedQuestion:∗∗<br>Ifthereare\displaystyle prizes is"<br><br>**1. Mathematical Impossibility of the Printed Question:**<br>If there are5$ students and each student is to receive $3\displaystyle prizes, the total number of prizes required would be:<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 prizes required</mtext><mo>=</mo><mn>5</mn><mo>×</mo><mn>3</mn><mo>=</mo><mn>15</mn><mtext> prizes</mtext></mrow><annotation encoding="application/x-tex">Total prizes required=5\displaystyle \text{Total prizes required} = 5 ×\displaystyle \times 3 = 15  prizes\displaystyle \text{ prizes}</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">Total prizes required</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:0.7278em;vertical-align:-0.0833em;"></span><span class="mord">5</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">3</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:0.8623em;vertical-align:-0.1944em;"></span><span class="mord">15</span><span class="mord text"><span class="mord"> prizes</span></span></span></span></span></span><br>However, we only have &#36;12 prizes. Therefore, it is mathematically impossible to divide $12prizesamong\displaystyle prizes among5$ students such that each receives $3\displaystyle prizes.<br><br>**2. Identifying and Resolving the Typographical Errors:**<br>This question contains multiple typographical errors in the textbook source. The mathematically correct formulation that leads to the textbook's key of **Option D** (&#36;16,800) is as follows:
- The total number of prizes is **$10\displaystyle ** (instead of &#36;12).
- The total number of students is **5\displaystyle 5**.
- The prizes are divided such that **three specific students receive 3\displaystyle 3 prizes each, one specific student receives 1\displaystyle 1 prize, and one specific student receives 0\displaystyle 0 prizes**.

Let us calculate the number of ways for this corrected distribution:
We are distributing $10$ distinct prizes into $5\displaystyle groups of sizes &#36;3, 3, 3, 1,  and \displaystyle \text{ and } 0 respectively. The number of ways to divide these distinct prizes among the 5\displaystyle 5 students is given by the multinomial coefficient:
Number of Ways=10!3!×3!×3!×1!×0!\displaystyle \text{Number of Ways} = \frac{10!}{3! ×\displaystyle \times 3! ×\displaystyle \times 3! ×\displaystyle \times 1! ×\displaystyle \times 0!}
Let us calculate this value:
10!=3,628,80010! = 3,628,800
3!×3!×3!×1!×0!=6×6×6×1×1=2163! ×\displaystyle \times 3! ×\displaystyle \times 3! ×\displaystyle \times 1! ×\displaystyle \times 0! = 6 ×\displaystyle \times 6 ×\displaystyle \times 6 ×\displaystyle \times 1 ×\displaystyle \times 1 = 216
Number of Ways=3,628,800216=16,800\displaystyle \text{Number of Ways} = \frac{3,628,800}{216} = 16,800
This matches **Option D** ($16,800$) exactly.

Hence, **Option D** 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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