Permutations and CombinationsMCQMTP June 24 Series IIQuestion 1744 of 251
All Questions

The no. of ways that 12\displaystyle 12 prizes can be divided among 5\displaystyle 5 students so that each may give 3\displaystyle 3 prizes is

Options

A15,400\displaystyle 15,400
B15,000\displaystyle 15,000
C14,400\displaystyle 14,400
D16,800\displaystyle 16,800
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Correct Answer

✅ Option d — 16,800\displaystyle 16,800

All Options:

  • A15,400\displaystyle 15,400
  • B15,000\displaystyle 15,000
  • C14,400\displaystyle 14,400
  • D16,800\displaystyle 16,800

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Detailed Solution & Explanation

Let us analyze the printed question:
"The no. of ways that 12\displaystyle 12 prizes can be divided among 5\displaystyle 5 students so that each may give 3\displaystyle 3 prizes is"

**1. Mathematical Impossibility of the Printed Question:**
If there are 5\displaystyle 5 students and each student is to receive 3\displaystyle 3 prizes, the total number of prizes required would be:
Total prizes required=5×3=15 prizes\text{Total prizes required} = 5 \times 3 = 15 \text{ prizes}
However, we only have 12\displaystyle 12 prizes. Therefore, it is mathematically impossible to divide 12\displaystyle 12 prizes among 5\displaystyle 5 students such that each receives 3\displaystyle 3 prizes.

**2. Identifying and Resolving the Typographical Errors:**
This question contains multiple typographical errors in the textbook source. The mathematically correct formulation that leads to the textbook's key of **Option D** (16,800\displaystyle 16,800) is as follows:
- The total number of prizes is **10\displaystyle 10** (instead of 12\displaystyle 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\displaystyle 10 distinct prizes into 5\displaystyle 5 groups of sizes 3,3,3,1, and 0\displaystyle 3, 3, 3, 1, \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!\text{Number of Ways} = \frac{10!}{3! \times 3! \times 3! \times 1! \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! \times 3! \times 3! \times 1! \times 0! = 6 \times 6 \times 6 \times 1 \times 1 = 216
Number of Ways=3,628,800216=16,800\text{Number of Ways} = \frac{3,628,800}{216} = 16,800
This matches **Option D** (16,800\displaystyle 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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