Permutations and CombinationsMCQMTP June 24 Series IQuestion 1675 of 251
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The letters of the word VIOLENT are arranged so that the vowels occupy even place only. The number of permutations is:

Options

A144
B120
C24
D72
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Correct Answer

✅ Option a — 144

All Options:

  • A144
  • B120
  • C24
  • D72

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Detailed Solution & Explanation

We want to find the number of permutations of the letters of the word "VIOLENT" such that the vowels occupy the even places only.

First, let us analyze the letters in the word "VIOLENT":
- Total number of letters = 7.
- Vowels: {I,O,E}\displaystyle \{I, O, E\} (3 vowels).
- Consonants: {V,L,N,T}\displaystyle \{V, L, N, T\} (4 consonants).

In a 7-letter arrangement, the positions are numbered 1,2,3,4,5,6,7\displaystyle 1, 2, 3, 4, 5, 6, 7.
- Even places: 2,4,6\displaystyle 2, 4, 6 (exactly 3 positions).
- Odd places: 1,3,5,7\displaystyle 1, 3, 5, 7 (exactly 4 positions).

Since there are exactly 3 vowels and exactly 3 even places, and the vowels must occupy the even places only, the 3 vowels must be arranged in these 3 even places:
Ways to arrange vowels=3!=3×2×1=6 ways\text{Ways to arrange vowels} = 3! = 3 \times 2 \times 1 = 6 \text{ ways}

The remaining 4 consonants must be arranged in the 4 odd places:
Ways to arrange consonants=4!=4×3×2×1=24 ways\text{Ways to arrange consonants} = 4! = 4 \times 3 \times 2 \times 1 = 24 \text{ ways}

Using the fundamental multiplication principle, the total number of valid permutations is:
Total permutations=3!×4!=6×24=144\text{Total permutations} = 3! \times 4! = 6 \times 24 = 144

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.

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Permutations vs Combinations: The Quantitative Aptitude Guide

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