document.write( "Question 1189433: What is the probability that a random arrangement of the letters in the word `SEVEN' will have both E's next to each other?
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Algebra.Com's Answer #820791 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Edit: I corrected the final answer to have it fully reduced. I appreciate @greenestamps pointing out the error.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If we could tell the two \"E\"s apart, then there would be 5! = 5*4*3*2*1 = 120 permutations of the word SEVEN since there are 5 letters.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "However, we cannot tell the \"E\"s apart. This means we've double-counted when computing that figure of 120. So we must cut that figure in half to get 60.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "There are 60 ways to arrange the letters of SEVEN where the \"E\"s may or may not be together.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's take the \"E\"s out and replace it with some other letter. I'll use W. \n" ); document.write( "So we go from the word SEVEN to the \"word\" SWVN\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "There are 4 distinct letters here, so there are 4! = 4*3*2*1 = 24 permutations. Anywhere you see a W, replace it with EE\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A permutation like WSVN is really EESVN \n" ); document.write( "A permutation like SVWN is really SVEEN \n" ); document.write( "and so on.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "--------------------------------------\r \n" ); document.write( "\n" ); document.write( "We found that there are:
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