Permutations and CombinationsMCQMTP Nov 20Question 1652 of 251
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How many ways can be letters of the word 'FAILURE' be arranged so that the consonants may occupy only odd places?

Options

A576
B476
C376
D276
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Correct Answer

Option a576

All Options:

  • A576
  • B476
  • C376
  • D276

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Detailed Solution & Explanation

The word is FAILURE, which contains 7 distinct letters: {F,A,I,L,U,R,E}\displaystyle \{F, A, I, L, U, R, E\}.
Let's separate them into vowels and consonants:
- Vowels: {A,I,U,E}\displaystyle \{A, I, U, E\} (4 vowels)
- Consonants: {F,L,R}\displaystyle \{F, L, R\} (3 consonants)
The 7 letter positions are:
1 (odd), 2 (even), 3 (odd), 4 (even), 5 (odd), 6 (even), 7 (odd)\text{1 (odd), 2 (even), 3 (odd), 4 (even), 5 (odd), 6 (even), 7 (odd)}
So, there are:
- 4 odd places: {1,3,5,7}\displaystyle \{1, 3, 5, 7\}
- 3 even places: {2,4,6}\displaystyle \{2, 4, 6\}
We want the 3 consonants to occupy *only* odd places.
1. Select 3 odd positions out of the 4 available odd positions and arrange the 3 consonants in them:
Ways=4P3=4×3×2=24 ways\text{Ways} = ^4P_3 = 4 \times 3 \times 2 = 24 \text{ ways}
2. The remaining 4 positions (3 even positions + 1 leftover odd position) must be filled by the 4 vowels:
Ways=4!=24 ways\text{Ways} = 4! = 24 \text{ ways}
Total number of arrangements is:
Total=4P3×4!=24×24=576\text{Total} = ^4P_3 \times 4! = 24 \times 24 = 576
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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