Permutations-Combinnations
Probability
Level
pending
All the letters of the word ‘EAMCOT’ are arranged in different possible ways. The number of such arrangements in which no two vowels are adjacent to each other is.
Direction - Consider the which are all places vowels could be placed placing the consonants.
36
576
144
720
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1 solution
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’.
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.