All the letters of the word ‘eamcot’ are arranged in different possible
ways. the number of such arrangements in which no two vowels are adjacent to each
other is
(a) 360 (b) 144 (c) 72 (d) 54

144 is correct.

Further Explanation:

Solution:

Number of arrangements in which no two vowels are adjacent to each other:

Given, word is ‘EAMCOT’

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:

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:

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

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

Subject: mathematics

Level: college

Keywords: solution, Vowels in word ‘EAMCOT’, formula.

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