Permutations and CombinationsPYQ May 25Question 4025 of 183
All Questions

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Options

A210
B1050
C25200
D21400
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option c — 25200

All Options:

  • A210
  • B1050
  • C25200
  • D21400

Detailed Solution & Explanation

To find the number of words of 3 consonants and 2 vowels that can be formed from 7 consonants and 4 vowels, we follow a two-step process:

1. **Selection of letters**:
- The number of ways to select 3 consonants from 7 is:
(73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35
- The number of ways to select 2 vowels from 4 is:
(42)=4×32×1=6\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
By the fundamental counting principle, the total number of ways to select these 5 letters is:
35×6=210 combinations35 \times 6 = 210 \text{ combinations}

2. **Arrangement of selected letters**:
Each selected set of 5 letters can be arranged among themselves to form a word in 5!\displaystyle 5! ways:
5!=5×4×3×2×1=120 ways5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \text{ ways}

3. **Total words formed**:
Total words=210×120=25200\text{Total words} = 210 \times 120 = 25200
Hence, **Option C** 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.

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

Start Practicing — It's Free