document.write( "Question 1159623: How many arrangements of the word ACTIVE are there if the C and E\r
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Algebra.Com's Answer #782757 by ikleyn(52778)\"\" \"About 
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document.write( "(a)  If the C and E are always together, we can consider this pair as one merged object.\r\n" );
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document.write( "     Then we have 5 objects in total that admit 5! = 1*2*3*4*5 = 120 permutations.\r\n" );
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document.write( "     But the merged object has two states: CE and EC.\r\n" );
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document.write( "     By combining all these permutations, we have, in total, 2*5! = 2*120 = 240 permutations.\r\n" );
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document.write( "     It is the ANSWER in this case.\r\n" );
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document.write( "(b)  This time, we should consider all permutations of 6 letters of the word ACTIVE, 6! = 720 permutations, \r\n" );
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document.write( "      and subtract those 2*120 = 240 permutations, found in the case (a), where the letters C and E are placed together.\r\n" );
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document.write( "      So, the ANSWER  in this case is  6! - 2*5! = 720 - 240 = 480.\r\n" );
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\n" ); document.write( "\n" ); document.write( "For many other similar solved problems, see the lesson\r
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\n" ); document.write( "\n" ); document.write( "Also,  you have this free of charge online textbook in ALGEBRA-II in this site\r
\n" ); document.write( "\n" ); document.write( "    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.\r
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\n" ); document.write( "\n" ); document.write( "The referred lesson is the part of this online textbook under the topic  \"Combinatorics: Combinations and permutations\". \r
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\n" ); document.write( "\n" ); document.write( "Save the link to this textbook together with its description\r
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\n" ); document.write( "\n" ); document.write( "Free of charge online textbook in ALGEBRA-II
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