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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