Question

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.

Answer 1

the ans is 4! * 3!. 

4! is the number of permutations keeping the vowels together. 
3! is the number of permutations of not keeping the vowels together. 

hence, 4! * 3! = 144. 

Answer 2

When all the letters of the word 'EAMCOT' are arranged in different possible ways are arranged then the answer will be 4! * 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! * 3! = 144.