document.write( "Question 1160357: How many permutations of 2 item can be selected from a group of 4? ?
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document.write( "B) Use the letters A, B, C D to identify the items & list each possibility. \r
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document.write( "c) If random sampling is employed what is the probability that any particular sample will be selected\r
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document.write( "is my answer correct below ? \r
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document.write( "(4!)/(2!) = 12\r
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document.write( "B) { AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC } \n" );
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Algebra.Com's Answer #783646 by ikleyn(52781)![]() ![]() You can put this solution on YOUR website! . \n" ); document.write( " \r\n" ); document.write( "\r\n" ); document.write( "The formula (4!)/(2!) = 12 produces the correct number/answer 12;\r\n" ); document.write( "\r\n" ); document.write( " but the logic behind it is DARK.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "The common/normal/regular logic is \r\n" ); document.write( "\r\n" ); document.write( " the number of permutations of 2 items selected from a group of 4 items is 4*3 = 12.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "It says that you can place any of 4 items to the 1-st position\r\n" ); document.write( "\r\n" ); document.write( "and then to complete the pair by adding any of 3 remaining items.\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |