Permutations and CombinationsMCQMTP March 21Question 1719 of 251
All Questions

An examination paper consists of 12 questions divided into two parts A and B. Part A contains 7 questions and Part B contains 5 questions. A candidate is required to attempt 8 questions selecting at least 3 from each part, in how many maximum ways can the candidate select the questions?

Options

A35
B175
C210
D420
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Correct Answer

✅ Option d — 420

All Options:

  • A35
  • B175
  • C210
  • D420

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Detailed Solution & Explanation

We are given:
- Total number of questions = 12\displaystyle 12
- Number of questions in Part A = 7\displaystyle 7
- Number of questions in Part B = 5\displaystyle 5
- Total questions to attempt = 8\displaystyle 8
- Constraint: The candidate must select **at least 3 questions** from each part.

Let us find the possible combinations of questions selected from Part A and Part B:
Since the total number of questions to attempt is 8\displaystyle 8, and each part must have at least 3\displaystyle 3 questions, the possible distributions (a,b)\displaystyle (a, b) where a\displaystyle a is from Part A and b\displaystyle b is from Part B are:
1. **Case 1:** 3\displaystyle 3 questions from Part A and 5\displaystyle 5 questions from Part B (3+5=8\displaystyle 3 + 5 = 8)
2. **Case 2:** 4\displaystyle 4 questions from Part A and 4\displaystyle 4 questions from Part B (4+4=8\displaystyle 4 + 4 = 8)
3. **Case 3:** 5\displaystyle 5 questions from Part A and 3\displaystyle 3 questions from Part B (5+3=8\displaystyle 5 + 3 = 8)

Let us calculate the number of ways for each case:

**Case 1: 3 from Part A and 5 from Part B**
Ways1=7C3×5C5\text{Ways}_1 = ^{7}C_{3} \times ^{5}C_{5}
7C3=7×6×53×2×1=35^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
5C5=1^{5}C_{5} = 1
Ways1=35×1=35\text{Ways}_1 = 35 \times 1 = 35

**Case 2: 4 from Part A and 4 from Part B**
Ways2=7C4×5C4\text{Ways}_2 = ^{7}C_{4} \times ^{5}C_{4}
7C4=7C3=35^{7}C_{4} = ^{7}C_{3} = 35
5C4=5^{5}C_{4} = 5
Ways2=35×5=175\text{Ways}_2 = 35 \times 5 = 175

**Case 3: 5 from Part A and 3 from Part B**
Ways3=7C5×5C3\text{Ways}_3 = ^{7}C_{5} \times ^{5}C_{3}
7C5=7×62×1=21^{7}C_{5} = \frac{7 \times 6}{2 \times 1} = 21
5C3=5×42×1=10^{5}C_{3} = \frac{5 \times 4}{2 \times 1} = 10
Ways3=21×10=210\text{Ways}_3 = 21 \times 10 = 210

**Total number of ways:**
Total Ways=Ways1+Ways2+Ways3\text{Total Ways} = \text{Ways}_1 + \text{Ways}_2 + \text{Ways}_3
Total Ways=35+175+210=420\text{Total Ways} = 35 + 175 + 210 = 420

**Discrepancy Note:**
The mathematical derivation establishes that the total number of ways is 420\displaystyle 420, which corresponds to **Option D**. The textbook answer key has a typographical error, listing **Option C** (210\displaystyle 210, which is only the number of ways for Case 3) as the correct answer.

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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