All the letters of the word 'EAMCET' are arranged in different possible ways.

Total number of words using letters of  EAMCET is

$\frac{6 !}{2}=\frac{6 \times 5 \times 4 \times 3 \times 1}{2}$ (Since $\mathrm{E}$ is respected twice)

= 360

Total number of words with no two vowels together is :-

Since vowels are $E, A, E=3$ and consonant is $M_{1} C_{1} T=3$.

i.e number of ways to arrange consonant is 3!

– C – C – C –

i.e vowel can take any of 4 places 

i.e number of vowels arrangement $=3 ! \times{ }^{4} P_{3} \times \frac{1}{2}(\because E$ is repeated twice $)$

i.e $3 ! \times \frac{4 !}{1 !} \times \frac{1}{2}=72$

$\therefore$ Number of words arrangement so that two vowels are adjacent $=\frac{360-72}{2}=144$