(B) is the correct choice.
We note that there are 3 consonants and 3
vowels E, A and O. Since no two vowels have to be together, the possible choice for
vowels are the places marked as ‘X’.
X M X C X T X, these vowels can be arranged
in 4P3 ways 3 consonants can be arranged in 3 ways.
Hence, the required number of
ways = 3! × 4P3
= 144
