Question 829874
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First we calculate the number of distinguishable permutations of EAMCET.
The 2 E's are not distinguishable so that number is 

     {{{6!/2!}}} = 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: {{{6!/2!}}}{{{""-""}}}{{{3*4!}}}{{{""-""}}}{{{3(5!-2*4!)}}} = {{{360 - 72 - 216 = 72}}}

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</pre>