Permutations and CombinationsMCQMTP Dec 2023 Series IIQuestion 1683 of 251
All Questions

A bag contains 4 red, 3 black and 2 white balls > In how many ways 3 balls can be drawn from this bag so that they include at least one black ball?

Options

A64
B31
C85
DNone of these
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Correct Answer

✅ Option a — 64

All Options:

  • A64
  • B31
  • C85
  • DNone of these

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Detailed Solution & Explanation

The bag contains:
- 4\displaystyle 4 Red balls
- 3\displaystyle 3 Black balls
- 2\displaystyle 2 White balls

Total number of balls in the bag:
Total Balls=4+3+2=9 balls\text{Total Balls} = 4 + 3 + 2 = 9 \text{ balls}
We need to draw 3\displaystyle 3 balls from the bag such that they include **at least one black ball**.
We can solve this using the subtraction method:
Favorable Ways=Total ways to choose 3 balls−Ways to choose 3 balls with NO black ball\text{Favorable Ways} = \text{Total ways to choose 3 balls} - \text{Ways to choose 3 balls with NO black ball}

- **Step 1: Calculate the total ways to choose any 3 balls from 9:**
Total Ways=9C3=9!3!×6!=9×8×73×2×1=84 ways\text{Total Ways} = ^{9}C_{3} = \frac{9!}{3! \times 6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \text{ ways}
- **Step 2: Calculate the ways to choose 3 balls with no black balls:**
If no black balls are chosen, all 3\displaystyle 3 balls must be chosen from the non-black balls (Red and White).
Total non-black balls=4 Red+2 White=6 balls\text{Total non-black balls} = 4 \text{ Red} + 2 \text{ White} = 6 \text{ balls}
Ways with no black ball=6C3=6!3!×3!=6×5×43×2×1=20 ways\text{Ways with no black ball} = ^{6}C_{3} = \frac{6!}{3! \times 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \text{ ways}
- **Step 3: Calculate the favorable ways:**
Favorable Ways=84−20=64 ways\text{Favorable Ways} = 84 - 20 = 64 \text{ ways}

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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