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