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