Permutations and CombinationsMCQMTP June 24 Series IIIQuestion 1746 of 251
All Questions

A box contains 7\displaystyle 7 red, 6\displaystyle 6 white and 4\displaystyle 4 blue balls. How many selections of three balls can be made so that none is red?

Options

A90\displaystyle 90
B120\displaystyle 120
C48\displaystyle 48
DNone of these
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — 120\displaystyle 120

All Options:

  • A90\displaystyle 90
  • B120\displaystyle 120
  • C48\displaystyle 48
  • DNone of these

Ad

Detailed Solution & Explanation

The box contains:
- 7\displaystyle 7 Red balls
- 6\displaystyle 6 White balls
- 4\displaystyle 4 Blue balls

We need to select 3\displaystyle 3 balls such that **none is red**.
This means all 3\displaystyle 3 selected balls must be chosen exclusively from the non-red balls (White and Blue balls).

Let us find the total number of non-red balls:
Non-red balls=6 White+4 Blue=10 balls\text{Non-red balls} = 6 \text{ White} + 4 \text{ Blue} = 10 \text{ balls}
The number of ways to select 3\displaystyle 3 balls from these 10\displaystyle 10 non-red balls is given by:
10C3=10!3!×(10−3)!^{10}C_{3} = \frac{10!}{3! \times (10-3)!}
10C3=10×9×83×2×1^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}
  ⟹  10C3=10×3×4=120\implies ^{10}C_{3} = 10 \times 3 \times 4 = 120

Hence, **Option B** 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

More Questions from Permutations and Combinations

The value of N\displaystyle N in N!+17!=18!\displaystyle N! + \frac{1}{7!} = \frac{1}{8!} is

A person can go from place 'A' to 'B' by 11 different modes of transport but is allowed to return to 'A' by any mode other than the one earlier. The number of different ways in which the entire journey can be completed is:

If a man travels from place A to B in 10 ways then by how many ways can he come back by another train?

If n!10(n−1)!=(n−1)!(n−3)!\displaystyle \frac{n!}{10(n-1)!} = \frac{(n-1)!}{(n-3)!} find 'n'.

Which of the following is a correct statement.

nP3=2⋅nP2\displaystyle ^nP_3 = 2 \cdot ^nP_2. Find n\displaystyle n.

Ready to Master Permutations and Combinations?

Practice all 251 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free