Permutations and CombinationsMCQPYQ Nov 20Question 1688 of 251
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Out of 7 boys and 4 girls, a team of a debate club of 5 is to be chosen. The number of teams such that each team includes at least one girl is:

Options

A439
B429
C419
D441
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Correct Answer

✅ Option d — 441

All Options:

  • A439
  • B429
  • C419
  • D441

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Detailed Solution & Explanation

We have a total of 7 boys+4 girls=11\displaystyle 7 \text{ boys} + 4 \text{ girls} = 11 children.
We want to choose a team of 5 children such that the team includes at least one girl. We can use the complement method.

Ways with at least one girl=Total ways to choose 5 children−Ways with no girls (all boys)\text{Ways with at least one girl} = \text{Total ways to choose 5 children} - \text{Ways with no girls (all boys)}

1. **Total ways to choose 5 children from 11:**
11C5=11×10×9×8×75×4×3×2×1=11×2×3×7=462 ways^{11}C_5 = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 11 \times 2 \times 3 \times 7 = 462 \text{ ways}

2. **Ways to choose 5 children with no girls (all 5 are boys):**
We select 5 boys from the 7 available boys:
7C5=7C2=7×62×1=21 ways^7C_5 = ^7C_2 = \frac{7 \times 6}{2 \times 1} = 21 \text{ ways}

3. **Ways with at least one girl:**
Ways=462−21=441\text{Ways} = 462 - 21 = 441

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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