Permutations and CombinationsMCQPYQ May 18Question 1718 of 249
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In an examination a candidate has to pass in 4 of the 4 papers. In how many different ways can be failed?

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

✅ Option c — 15

All Options:

  • A14
  • B16
  • C15
  • DNone

Detailed Solution & Explanation

For each of the 4\displaystyle 4 papers, a candidate has 2\displaystyle 2 possible outcomes: either **Pass (P)** or **Fail (F)**.

Therefore, the total number of possible outcomes for all 4\displaystyle 4 papers is:
Total Outcomes=2×2×2×2=24=16Total Outcomes=2\displaystyle \text{Total Outcomes} = 2 ×\displaystyle \times 2 ×\displaystyle \times 2 ×\displaystyle \times 2 = 2^4 = 16
A candidate is considered to have successfully passed the examination if and only if they pass in all 4\displaystyle 4 papers. There is exactly 1\displaystyle 1 way to achieve this outcome:
Ways to Pass=1(i.e., PPPP)Ways to Pass=1\displaystyle \text{Ways to Pass} = 1 \quad (i.e., \displaystyle \text{i.e., } PPPP)
A candidate fails the examination if they fail in one or more papers. Thus, the number of ways to fail is all outcomes except the one where the candidate passes all papers:
Ways to Fail=Total Outcomes−Ways to Pass\displaystyle \text{Ways to Fail} = \text{Total Outcomes} - Ways to Pass\displaystyle \text{Ways to Pass}
Ways to Fail=16−1=15Ways to Fail=16−1=15\displaystyle \text{Ways to Fail} = 16 - 1 = 15
Alternatively, we can sum the number of ways to fail in exactly 1\displaystyle 1, 2\displaystyle 2, 3\displaystyle 3, or 4\displaystyle 4 papers:
- Fail in 1\displaystyle 1 paper: 4C1=4\displaystyle ^{4}C_{1} = 4 ways
- Fail in 2\displaystyle 2 papers: 4C2=6\displaystyle ^{4}C_{2} = 6 ways
- Fail in 3\displaystyle 3 papers: 4C3=4\displaystyle ^{4}C_{3} = 4 ways
- Fail in 4\displaystyle 4 papers: 4C4=1\displaystyle ^{4}C_{4} = 1 way
Total Ways to Fail=4+6+4+1=15Total Ways to Fail=4+6+4+1=15\displaystyle \text{Total Ways to Fail} = 4 + 6 + 4 + 1 = 15
**Discrepancy Note:**
The mathematical solution clearly shows there are $15\displaystyle ways to fail, which matches **Option C**. The textbook answer key has a typographical error, indicating **Option A** ($14) as the correct answer. This error sometimes occurs in prep materials due to an incorrect formula (like subtracting 2\displaystyle 2 instead of 1\displaystyle 1).

Hence, **Option C** 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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