Q. 32

# All the letters of the word ‘EAMCOT’ are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.

Given, word is ‘EAMCOT’

To find: number of arrangements in which no two vowels are adjacent to each other

Vowels in word ‘EAMCOT’ = 3(E, A, O) and consonants = 3(M, C, T)

Let vowels be denoted by V

Now, fix the position by Vowels like this:

The remaining 4 places can be occupied by 3 consonants

Now, arrange 3 consonants at 4 places and 3 vowels at 3 places

Formula used:

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

Total number of arrangements of vowels

= the number of arrangements of 3 things taken all at a time

= P(3, 3)

{ 0! = 1}

= 3!

= 3 × 2 × 1

= 6

Formula used:

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

Total number of arrangements of consonants

= the number of arrangements of 4 things taken 3 at a time

= P(4, 3)

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, total number of arrangements in which no two vowels are adjacent to each other, 6 × 24 = 144

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