Permutations and CombinationsMCQPYQ Sep 24Question 1704 of 251
All Questions

How many total combinations can be formed of 8 different counters marked as 1, 2, 3, 4, 5, 6, 7, and 8, taking 4 counters at a time and there being at least one odd and one even numbered counter in each combination?

Options

A66
B68
C64
D62
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — 68

All Options:

  • A66
  • B68
  • C64
  • D62

Ad

Detailed Solution & Explanation

We have 8 counters marked with numbers {1,2,3,4,5,6,7,8}\displaystyle \{1, 2, 3, 4, 5, 6, 7, 8\}.
Let's classify them into odd and even numbers:
- Odd counters: {1,3,5,7}\displaystyle \{1, 3, 5, 7\} (4 counters)
- Even counters: {2,4,6,8}\displaystyle \{2, 4, 6, 8\} (4 counters)
We want to select 4 counters such that there is *at least one odd* and *at least one even* counter in the combination.
1. **Total ways** to choose 4 counters from 8 without any restriction:
Total=8C4=8×7×6×54×3×2×1=70 ways\text{Total} = ^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70 \text{ ways}
2. **Unwanted combinations**:
- **All even counters**: Choosing 4 even counters out of 4:
Ways=4C4=1 way\text{Ways} = ^4C_4 = 1 \text{ way}
- **All odd counters**: Choosing 4 odd counters out of 4:
Ways=4C4=1 way\text{Ways} = ^4C_4 = 1 \text{ way}
3. **Valid combinations** (at least one of each):
Valid=Total−(All even+All odd)=70−(1+1)=68\text{Valid} = \text{Total} - (\text{All even} + \text{All odd}) = 70 - (1 + 1) = 68
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.

Key Concepts to Understand

Combination

A selection of objects where order does not matter.

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