document.write( "Question 360174: Please help me solve this:
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document.write( "Jury selection. In how many ways can 12 jurors and 2 alternates be chosen from a group of 20 prospective jurors? \n" );
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Algebra.Com's Answer #256992 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! 14 are to be chosen. \n" ); document.write( "The 1st is 1 of 20. \n" ); document.write( "The 2nd is 1 of 19, etc \n" ); document.write( "--> 20*19*18*17*16*15*14*13*12*11*10*9*8*7...1 \n" ); document.write( "-------------- \n" ); document.write( "But, since a jury of A, B, C, D ... etc is the same as A, D, B, C... it's necessary to divide by 14*13*12*...1 \n" ); document.write( "The result is 20*19*18*17*16*15*14*13*12*11*10*9*8*7 \n" ); document.write( "This is 20!/((20-14)!*14!) \n" ); document.write( "= 20!/14!6! \n" ); document.write( "= 38760 possibilities \n" ); document.write( "--------------\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |