All the letters of the word 'EAMCOT' are arranged in different possible ways.

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,

${ }_{-} \mathrm{M}_{-} \mathrm{C}_{-} \mathrm{T}_{-}$

These vowels can be arranged in 4P3 ways.

3 consonants can be arranged in 3! ways.

Hence, the required numbers of ways = 3! × 4P3 = 144 ways.