Permutations and CombinationsMCQPYQ Jun 19Question 1686 of 249
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In how many ways can 4 people be selected at random from 6 boys and 4 girls if there are to be exactly 2 girls?

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

✅ Option a — 90

All Options:

  • A90
  • B360
  • C92
  • D480

Detailed Solution & Explanation

We want to select 4 people from a group of 6 boys and 4 girls such that the selection contains exactly 2 girls.

Since the total selection must consist of 4 people and exactly 2 must be girls, the remaining 4−2=2\displaystyle 4 - 2 = 2 people must be boys.

1. **Selecting the Girls:**
We select 2 girls from the 4 available girls. The number of ways to do this is:
4C2=4×32×1=6 ways^4C_2 = 4×32×1=6\displaystyle \frac{4 \times 3}{2 \times 1} = 6  ways\displaystyle \text{ ways}

2. **Selecting the Boys:**
We select 2 boys from the 6 available boys. The number of ways to do this is:
6C2=6×52×1=15 ways^6C_2 = 6×52×1=15\displaystyle \frac{6 \times 5}{2 \times 1} = 15  ways\displaystyle \text{ ways}

Using the fundamental multiplication principle, the total number of ways to make the selection is:
Total ways=4C2×6C2=6×15=90Total ways=4C2\displaystyle \text{Total ways} = ^4C_2 ×\displaystyle \times ^6C_2 = 6 ×\displaystyle \times 15 = 90

Hence, **Option A** 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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