How many permutations can be formed by the letters of the word vowels when all consonants come together?

Book: RD Sharma - Mathematics

Chapter: 16. Permutations

Subject: Maths - Class 11th

Q. No. 7 of Exercise 16.4

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7

How many permutations can be formed by the letters of the word, ‘VOWELS,’ when

ach word begins with O and ends with L?

Given: the word is ‘VOWELS.’

To find: number of words using letters of the given word starting with O and ending with L

So, fix the position of first and last letter as O and L:

Remaining number of letters = 4

Now, we need to arrange these 4 letters at the remaining 4 places.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, the possible number of words using letters of ‘VOWELS’ starting with ‘O’ and ending with ‘L’ is 24

More Exercise Questions

How many permutations can be formed by the letters of the word, ‘VOWELS’, when

Question:

How many permutations can be formed by the letters of the word, ‘VOWELS’, when

(i) there is no restriction on letters?

(ii) each word begins with E?

(iii) each word begins with O and ends with L?

(iv) all vowels come together?

(v) all consonants come together?

Solution:

(i) The word VOWELS consists of 6 distinct letters that can be arranged amongst themselves in 6! ways.

∴ Number of words that can be formed with the letters of the word VOWELS, without any restriction = 6! = 720

(ii) If we fix the first letter as E, the remaining 5 letters can be arranged in 5! ways to form the words.

∴ Number of words starting with the E = 5! = 120

(iii) If we fix the first letter as O and the last letter as L, the remaining 4 letters can be arranged in 4! ways to form the words.

∴  Number of words that start with O and end with L = 4! = 24

(iv) The word VOWELS consists of 2 vowels.

If we keep all the vowels together, we have to consider them as a single entity.

Now, we are left with the 4 consonants and all the vowels that are taken together as a single entity.

This gives us a total of 5 entities that can be arranged in 5! ways.

It is also to be considered that the 2 vowels can be arranged in 2! ways amongst themselves.

By fundamental principle of counting:

$\therefore$ Total number of arrangements $=5 ! \times 2 !=240$

(v) The word VOWELS consists of 4 consonants.

If we keep all the consonants together, we have to consider them as a single entity.

Now, we are left with the 2 vowels and all the consonants that are taken together as a single entity.

This gives us a total of 3 entities that can be arranged in 3! ways.

It is also to be considered that the 4 consonants can be arranged in 4! ways amongst themselves.

By fundamental principle of counting:

$\therefore$ Total number of arrangements $=3 ! \times 4 !=144$

How many permutations can be formed by the letters of the word, 'VOWELS', when

all consonants come together?

The word VOWELS consists of 4 consonants.
If we keep all the consonants together, we have to consider them as a single entity.
Now, we are left with the 2 vowels and all the consonants that are taken together as a single entity.
This gives us a total of 3 entities that can be arranged in 3! ways.
It is also to be considered that the 4 consonants can be arranged in 4! ways amongst themselves.

By fundamental principle of counting:
∴ Total number of arrangements = 3!\[\times\]4! = 144

Concept: Factorial N (N!) Permutations and Combinations

  Is there an error in this question or solution?

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