document.write( "Question 500253: You have the word FEEBLENESS, in a random arrangement, what is the probabilty that exactly three E's will be together? \n" ); document.write( "
Algebra.Com's Answer #338102 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "FEEBLENESS\r\n" );
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document.write( "To have three E's together we have to find the number of all cases of \r\n" );
document.write( "distinguishable arrangements of these 8 things \r\n" );
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document.write( "B,E,F,L,N,S,S,(EEE)\r\n" );
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document.write( "and divide it by the number of all cases of distinguishable \r\n" );
document.write( "arrangements of these 10 things:\r\n" );
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document.write( "B,E,E,E,E,F,L,N,S,S,\r\n" );
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document.write( "The answer is \"8%21%2F10%21\" = \"%288%2A7%2A6%2A5%2A4%2A3%2A2%2A1%29%2F%2810%2A9%2A8%2A7%2A6%2A5%2A4%2A3%2A2%2A1%29\" =  = \"1%2F%2810%2A9%29\" = \"1%2F90\"\r\n" );
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document.write( "However you may think it's not that because there are indistinguishable letters E and S.  But we must divide\r\n" );
document.write( "both numerator and deminator by the same number to \"unorder\" them.   \r\n" );
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document.write( "To explain that:\r\n" );
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document.write( "Let's start out making the E's and S's distinguishable\r\n" );
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document.write( "B, E1, E2, E3, E4, F, L, N, S1, S2\r\n" );
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document.write( "There would be 10! ways to arrange those if we could tell the difference between the E's and the S's.\r\n" );
document.write( "But we can't.  So 10! counts this too many times.  We need to find out how many times too many 10! \r\n" );
document.write( "counts any arrangement.\r\n" );
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document.write( "To find that, look at an arbitrary arranglement of those 10 distinguishable things, say this one:\r\n" );
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document.write( "LE4FS2BE2NS1E3E1 \r\n" );
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document.write( "There are 4! ways the 4 subscripts of E could be arranged, and for each of those ways, there are 2!\r\n" );
document.write( "ways the subscripts of S could be arranged.  So the number of times too many which 10! counts the \r\n" );
document.write( "arrangement of LEFSBENSEE is 4!2!\r\n" );
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document.write( "However, any arbitrary distinguishable arrangement of the 8! when the E's\r\n" );
document.write( "are together, say,\r\n" );
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document.write( "S2L(E4E1E3)FBE2S1L\r\n" );
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document.write( "is also counted 4!2! times too many among the 8!, so we would divide it by the same number to\r\n" );
document.write( "\"unorder\" it.  So technically we would do it this way:\r\n" );
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document.write( "\"%288%21%2F%284%212%21%29%29%2F%2810%21%2F%284%212%21%29%29\"\"%22%22=%22%22\"\"expr%288%21%2F%284%212%21%29%29expr%28%284%212%21%29%2F10%21%29\"\"%22%22=%22%22\"\"expr%288%21%2F%28cross%284%212%21%29%29%29expr%28%28cross%284%212%21%29%29%2F10%21%29\"\"%22%22=%22%22\"\"8%21%2F10%21\"\"1%2F90\"\r\n" );
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document.write( "Edwin

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