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  4. 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

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Q. 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

Permutations and Combinations

Solution:

We note that there are $3$ consonants and $3$ vowels $E, A$ and $O$. Since no two vowels have to be together, the possible choice for vowels are the places marked as $ 'X'$ in $XMXCXTX$, these vowels can be arranged in $^{4}P_{3}$ ways, $3$ consonants can be arranged in $ \lfloor 3$ ways. Hence, the required number of ways $ = 3! \times ^{4}P_{3} = 3! \times 4! = 144$.