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.
Solution:
We note that, there are 3 consonants M, C, T and 3 vowels E, A, O.
Since, no two vowels have to be together, the possible choice for volwels are the blank spaces,
${ }_{-} \mathrm{M}_{-} \mathrm{C}_{-} \mathrm{T}_{-}$
These vowels can be arranged in 4P3 ways.
3 consonants can be arranged in 3! ways.
Hence, the required numbers of ways = 3! × 4P3 = 144 ways.