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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Solution
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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