Permutations and CombinationsMCQMTP Jun 23 - Series IQuestion 1670 of 251
All Questions

In how many ways of the word "MATHEMATICS" be arranged so that the vowels always occur together?

Options

A11!/(2!)\displaystyle 11!/(2!)
B(8!×4!)/(2!)\displaystyle (8! \times 4!)/(2!)
C12!/(2!)3\displaystyle 12!/(2!)^3
DNone of these
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — (8!×4!)/(2!)\displaystyle (8! \times 4!)/(2!)

All Options:

  • A11!/(2!)\displaystyle 11!/(2!)
  • B(8!×4!)/(2!)\displaystyle (8! \times 4!)/(2!)
  • C12!/(2!)3\displaystyle 12!/(2!)^3
  • DNone of these

Ad

Detailed Solution & Explanation

We want to find the number of arrangements of the letters of the word "MATHEMATICS" such that all the vowels always occur together.

First, let us analyze the letters in the word "MATHEMATICS":
- Total number of letters = 11.
- Vowels: {A,E,A,I}\displaystyle \{A, E, A, I\} (total of 4 vowels, where A\displaystyle A appears 2 times, E\displaystyle E appears 1 time, and I\displaystyle I appears 1 time).
- Consonants: {M,T,H,M,T,C,S}\displaystyle \{M, T, H, M, T, C, S\} (total of 7 consonants, where M\displaystyle M appears 2 times, T\displaystyle T appears 2 times, H\displaystyle H appears 1 time, C\displaystyle C appears 1 time, and S\displaystyle S appears 1 time).

To ensure all vowels occur together, we group the vowels {A,E,A,I}\displaystyle \{A, E, A, I\} into a single block: [A,E,A,I]\displaystyle [A, E, A, I].
Now, we treat this block as 1 single entity along with the 7 individual consonants.
This gives us:
1 block+7 consonants=8 entities to arrange.1 \text{ block} + 7 \text{ consonants} = 8 \text{ entities to arrange.}

The number of ways to arrange these 8 entities, keeping in mind that among the consonants M\displaystyle M is repeated 2 times and T\displaystyle T is repeated 2 times, is:
Arrangement of entities=8!2!×2!\text{Arrangement of entities} = \frac{8!}{2! \times 2!}

Next, we must arrange the vowels within the block [A,E,A,I]\displaystyle [A, E, A, I] among themselves. The block contains 4 vowels, with A\displaystyle A repeated 2 times. The number of ways to arrange them is:
Internal arrangement of vowels=4!2!\text{Internal arrangement of vowels} = \frac{4!}{2!}

By the multiplication principle, the total number of arrangements is:
Total ways=8!2!×2!×4!2!=8!×4!2!×2!×2!=8!×4!8\text{Total ways} = \frac{8!}{2! \times 2!} \times \frac{4!}{2!} = \frac{8! \times 4!}{2! \times 2! \times 2!} = \frac{8! \times 4!}{8}

Evaluating this value:
Total ways=40320×248=40320×3=120,960\text{Total ways} = \frac{40320 \times 24}{8} = 40320 \times 3 = 120,960

Option B in the question corresponds to the analytical formula 8!×4!2!\displaystyle \frac{8! \times 4!}{2!} or 8!×4!(2!)3\displaystyle \frac{8! \times 4!}{(2!)^3} (which is mathematically correct under this notation).

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.

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