Permutations and CombinationsMCQMTP Nov 19Question 1650 of 251
All Questions

In how many ways can the letters of the word 'STRANGE' be arranged so that the vowels never come together?

Options

A3600
B3686
C5040
D4050
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Correct Answer

✅ Option a — 3600

All Options:

  • A3600
  • B3686
  • C5040
  • D4050

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Detailed Solution & Explanation

The word is STRANGE, which contains 7 distinct letters: {S,T,R,A,N,G,E}\displaystyle \{S, T, R, A, N, G, E\}.
Let's separate the vowels and consonants:
- Vowels: {A,E}\displaystyle \{A, E\} (2 vowels)
- Consonants: {S,T,R,N,G}\displaystyle \{S, T, R, N, G\} (5 consonants)

1. **Total arrangements** of the 7 letters:
Total=7!=5040\text{Total} = 7! = 5040
2. **Arrangements where the vowels are always together**:
Treat the two vowels {A,E}\displaystyle \{A, E\} as a single block/entity (AE)\displaystyle (AE).
Then we have 6 entities to arrange: (AE)\displaystyle (AE), S\displaystyle S, T\displaystyle T, R\displaystyle R, N\displaystyle N, G\displaystyle G.
Number of ways to arrange these 6 entities = 6!=720\displaystyle 6! = 720.
Within the block, the 2 vowels can be arranged in 2!=2\displaystyle 2! = 2 ways (AE\displaystyle AE or EA\displaystyle EA).
So, the number of arrangements where vowels are together is:
Vowels together=720×2=1440\text{Vowels together} = 720 \times 2 = 1440
3. **Arrangements where the vowels are never together**:
Vowels never together=Total−Vowels together=5040−1440=3600\text{Vowels never together} = \text{Total} - \text{Vowels together} = 5040 - 1440 = 3600
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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