Permutations and CombinationsMCQMTP Sep 24 Series IIQuestion 1754 of 251
All Questions

A candidate is required to answer 6\displaystyle 6 out of 10\displaystyle 10 questions, which are divided into two groups each containing 5\displaystyle 5 questions and he is not permitted to attempt more than 4\displaystyle 4 from each group. In how many ways can he make up his choice?

Options

A315\displaystyle 315
B250\displaystyle 250
C450\displaystyle 450
D200\displaystyle 200
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Correct Answer

✅ Option d — 200\displaystyle 200

All Options:

  • A315\displaystyle 315
  • B250\displaystyle 250
  • C450\displaystyle 450
  • D200\displaystyle 200

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Detailed Solution & Explanation

We are given:
- Total number of questions = 10\displaystyle 10
- Group 1 has 5\displaystyle 5 questions.
- Group 2 has 5\displaystyle 5 questions.
- The candidate must answer exactly 6\displaystyle 6 questions in total.
- Constraint: The candidate cannot answer more than 4\displaystyle 4 questions from any single group.

Let g1\displaystyle g_1 be the number of questions attempted from Group 1, and g2\displaystyle g_2 be the number of questions attempted from Group 2. We must have:
g1+g2=6where g1≤4 and g2≤4g_1 + g_2 = 6 \quad \text{where } g_1 \le 4 \text{ and } g_2 \le 4
The possible valid combinations (g1,g2)\displaystyle (g_1, g_2) are:
1. **Case 1:** 2\displaystyle 2 questions from Group 1 and 4\displaystyle 4 questions from Group 2
2. **Case 2:** 3\displaystyle 3 questions from Group 1 and 3\displaystyle 3 questions from Group 2
3. **Case 3:** 4\displaystyle 4 questions from Group 1 and 2\displaystyle 2 questions from Group 2

Let us calculate the number of ways for each case:

- **Case 1: 2 from Group 1, 4 from Group 2**
Ways1=5C2×5C4=(5×42×1)×5=10×5=50\text{Ways}_1 = ^{5}C_{2} \times ^{5}C_{4} = \left(\frac{5 \times 4}{2 \times 1}\right) \times 5 = 10 \times 5 = 50
- **Case 2: 3 from Group 1, 3 from Group 2**
Ways2=5C3×5C3=10×10=100\text{Ways}_2 = ^{5}C_{3} \times ^{5}C_{3} = 10 \times 10 = 100
- **Case 3: 4 from Group 1, 2 from Group 2**
Ways3=5C4×5C2=5×10=50\text{Ways}_3 = ^{5}C_{4} \times ^{5}C_{2} = 5 \times 10 = 50
By the addition principle, the total number of ways the candidate can choose the questions is:
Total Ways=Ways1+Ways2+Ways3\text{Total Ways} = \text{Ways}_1 + \text{Ways}_2 + \text{Ways}_3
Total Ways=50+100+50=200\text{Total Ways} = 50 + 100 + 50 = 200

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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