SOLUTION: All letters of the word "EAMCET"are arranged in all possible ways.The number of such arrangement in which two vowels are not adjacent to each other.

First we calculate the number of distinguishable permutations of EAMCET.
The 2 E's are not distinguishable so that number is 

      = 360

Then from those 360 we will subtract 

(1) the number of ways all 3 vowels come together.
and
(2) the number of ways exactly 2 vowels come together.

We calculate (1), the number of ways all 3 vowels come together

I will use the notation N{i,j,k,...} to indicate the number of permutations
of the elements between the braces.

N{EEA,M,C,T} = N(EAE,M,C,T} = N{AEE,M,C,T} = 4! = 24

So the number of ways all three vowels can come together is 3·4! = 72

Next we calculate (2), the number of ways exactly 2 vowels come together

N{AE,E,M,C,T} - N{AEE,M,C,T} - N{EAE,M,C,T} = 
N{EA,E,M,C,T} - N{EAE,M,C,T} - N{EEA,M,C,T} = 
N{EE,A,M,C,T} - N{EEA,M,C,T} - N{AEE,M,C,T} = 5!-4!-4! = 5!-2*4! = 72

So the number of ways exactly 2 vowels come together = 3(5!-2*4!) = 216

Answer:  = 

As a check, with the aid of my computer, here are all 72 such permutations
in 8 rows of 9 each:

EMECAT, EMECTA, EMETAC, EMETCA, EMACET, EMACTE, EMATEC, EMATCE, EMCETA,
EMCATE, EMTECA, EMTACE, ECEMAT, ECEMTA, ECETAM, ECETMA, ECAMET, ECAMTE,
ECATEM, ECATME, ECMETA, ECMATE, ECTEMA, ECTAME, ETEMAC, ETEMCA, ETECAM,
ETECMA, ETAMEC, ETAMCE, ETACEM, ETACME, ETMECA, ETMACE, ETCEMA, ETCAME,
AMECET, AMECTE, AMETEC, AMETCE, AMCETE, AMTECE, ACEMET, ACEMTE, ACETEM,
ACETME, ACMETE, ACTEME, ATEMEC, ATEMCE, ATECEM, ATECME, ATMECE, ATCEME,
MECETA, MECATE, METECA, METACE, MACETE, MATECE, CEMETA, CEMATE, CETEMA,
CETAME, CAMETE, CATEME, TEMECA, TEMACE, TECEMA, TECAME, TAMECE, TACEME.

Edwin