document.write( "Question 360174: Please help me solve this:
\n" ); document.write( "Jury selection. In how many ways can 12 jurors and 2 alternates be chosen from a group of 20 prospective jurors?
\n" ); document.write( "

Algebra.Com's Answer #256992 by Alan3354(69443)\"\" \"About 
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( "
\n" );