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 $=3 ! \times 4 !=144$