document.write( "Question 467690: how many ways can you arrange A-F with a consonant at one end, and a vowel at the other?\r
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document.write( "we answered other similiar problems using factorials \n" );
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Algebra.Com's Answer #320925 by ccs2011(207)![]() ![]() You can put this solution on YOUR website! **For any set of n objects, there are n! ways of arranging them** \n" ); document.write( "Now for the letters A-F, there are 2 vowels and 4 consonants \n" ); document.write( "If a vowel is at one end and a consonant at the other then that leaves 4 letters remaining to be arranged in any which way between them. \n" ); document.write( "From above we know then that are 4! ways of arranging 4 letters. \n" ); document.write( "Say the consonant B is on the front end and a vowel on other end: \n" ); document.write( "B _ _ _ _ Vowel \n" ); document.write( "There are only 2 possibilities because there are only 2 vowels to choose from: \n" ); document.write( "B _ _ _ _ A OR B _ _ _ _ E \n" ); document.write( "Each of these has 4! arrangements \n" ); document.write( "So with B at the front there are 2*(4!) arrangements \n" ); document.write( "However, there are 4 consonants each of which could be at the front end \n" ); document.write( "So multiply by 4 \n" ); document.write( "With any consonant at front end there are 4*2*(4!) arrangements \n" ); document.write( "However, the consonant could also be at back end with vowel in front \n" ); document.write( "Since everything else would stay the same, double the possible arrangements \n" ); document.write( "Total number of arrangements are 2*4*2*(4!) arrangements which simplifies to 16(4!) \n" ); document.write( " |