All the letters of the word EAMCOT are arranged in a different possible ways. Find the number of arrangments in which no two vowels are adjacent to each other.

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EMCOT
Vowels - A,E,O
number of vowels = 3
arrangement of vowels = 3! =1 x 2 x 3 = 6
number of arrangement of other letters = 4! = 1 x 2 x 3 x 4 = 24
Total number of arrangements = 3! x 4! =6 x 24 = 144