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The number of permutation from the letter A to G, so that neither the set BEG nor CAD appears is ?
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document.write( "The number of ways that ABCDEFG can be arranged is 7!\r\n" );
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document.write( "From that we must subtract the number of ways BEG and/or CAD appear.\r\n" );
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document.write( "We use\r\n" );
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document.write( "N(X or Y) = N(X) + N(Y) - N(X and Y)\r\n" );
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document.write( "Where \r\n" );
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document.write( "N(X) = \"The number of ways in which BEG appears\",\r\n" );
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document.write( "N(Y) = \"The number of ways in which CAD appears\"\r\n" );
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document.write( "N(X and Y) = \"The number of ways in which both BED and CAD appear\"\r\n" );
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document.write( "To find N(X):\r\n" );
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document.write( "The number of ways BEG occurs is in one of these 5 configurations:\r\n" );
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document.write( "BEG----, -BEG---, --BEG--, ---BEG-, ----BEG\r\n" );
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document.write( "where the other letters {A,C,D,F} are where the dashes are.\r\n" );
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document.write( "We can choose the configuration for BEG to appear in 5 ways.\r\n" );
document.write( "We choose the arrangements of the other 4 letters to go where\r\n" );
document.write( "the dashes are in 4! or 24 ways.\r\n" );
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document.write( "So N(X) = 5×4! = 5×24 = 120\r\n" );
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document.write( "To find N(Y)\r\n" );
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document.write( "N(Y) is the same and N(X), so N(Y) = 120 ways also.\r\n" );
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document.write( "To find N(X and Y)\r\n" );
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document.write( "The number of ways both BEG and CAD occur together in one of \r\n" );
document.write( "these 6 configurations:\r\n" );
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document.write( "BEGCADF, BEGFCAD, FBEGCAD, CADBEGF, CADFBEG, FCADBEG\r\n" );
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document.write( "So N(X and Y) = 6\r\n" );
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document.write( "N(X or Y) = N(X) + N(Y) - N(X and Y)\r\n" );
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document.write( " = 120 + 120 - 6 = 234\r\n" );
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document.write( "Answer: 7! - 234 = 5040 - 234 = 4806\r\n" );
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document.write( "Edwin \n" ); document.write( " \n" ); document.write( "