document.write( "Question 1194881: How many words can be made from the word \"EAMCET\" if we take two vowels together? \n" ); document.write( "
Algebra.Com's Answer #827201 by Edwin McCravy(20055)\"\" \"About 
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document.write( "In \"EAMCET\", there are 3 vowels and 3 consonants.  The consonants are all\r\n" );
document.write( "distinguishable, but two of the vowels, the 2 E's, are indistinguishable. \r\n" );
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document.write( "First we calculate the total number of arrangements with no restrictions \r\n" );
document.write( "other than that all the arrangements be distinguishable.  That number would \r\n" );
document.write( "be 6! if all letters were distinguishable, but we must divide by 2! because\r\n" );
document.write( "of 2 indistinguishable E's.  So if there were no restrictions other than\r\n" );
document.write( "that they be indistinguishable, the answer would be 6!/2! = 720/2 = 360\r\n" );
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document.write( "However, there are 4 types of arrangements that are not allowed,\r\n" );
document.write( "those in which no vowels are together. The number of these must be \r\n" );
document.write( "subtracted from the 360. These types are:\r\n" );
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document.write( "1. VCVCVC\r\n" );
document.write( "2. VCVCCV\r\n" );
document.write( "3. VCCVCV\r\n" );
document.write( "4. CVCVCV\r\n" );
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document.write( "where C stands for a consonant and V stands for a vowel. \r\n" );
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document.write( "For each of those 4 types, there are 3!=6 ways to arrange \r\n" );
document.write( "the 3 consonants.\r\n" );
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document.write( "For each of those (4)(6)=24 ways to pick the types and \r\n" );
document.write( "arrangements of the 3 consonants, there are 3!/2!=6/2=3 \r\n" );
document.write( "ways to choose the vowels. We divided by 2! because\r\n" );
document.write( "there are 2 indistinguishable vowels, the E's.\r\n" );
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document.write( "So there are (4)(6)(3)=72 arrangements that are not allowed.\r\n" );
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document.write( "Answer: 360 - 72 = 288 ways\r\n" );
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document.write( "Edwin
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