All the letters of the word $^{'} E R M C O T^{'}$ are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.

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Solution

When all the letters of the word 'EAMCOT' are arranged in different possible ways are arranged then the answer will be $4! \times 3!$ .

$3!$ is the number of permutations of not keeping vowels together in the word.

$4!$ is the number of permutations keeping vowels together in the word.
Therefore the correct answer will be $4! \times 3! = 144$ .

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