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

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 $=3 ! \times 4 !=144$