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.
We note that, there are 3 consonants M, C, T and 3 vowels E, A, O.
Since, no two vowels have to be together, the possible choice for volwels are the blank spaces,
These vowels can be arranged in P ways.
3 consonants can be arranged in 3! ways.
Hence, the required numbers of ways = 3! × P = 144 ways.