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.

Answer :

**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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Which of the following are true :

(2 + 3)! = 2! + 3!

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