document.write( "Question 1046393: A parade is to include five carriages A, B, C, D and E. In how many ways can the carriages be ordered if B must not precede D? \n" ); document.write( "
Algebra.Com's Answer #661863 by t0hierry(194)\"\" \"About 
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3! counts how many ways you can order A,C,E. 4! will count how many ways you can order A C E D.
\n" ); document.write( "ACE D
\n" ); document.write( "AEC D
\n" ); document.write( "CEA D
\n" ); document.write( "CAE D
\n" ); document.write( "EAC D
\n" ); document.write( "ECA D
\n" ); document.write( "AC D E
\n" ); document.write( "AE D C
\n" ); document.write( "CE D A
\n" ); document.write( "CA D E
\n" ); document.write( "EA D C
\n" ); document.write( "EC D A\r
\n" ); document.write( "\n" ); document.write( "Repeat with D in second position, and in first.
\n" ); document.write( "For D in 4th position, B is imposed. So we have 6 ways.
\n" ); document.write( "For D in 3rd position, we have 2 possibilities for each B (2*6)
\n" ); document.write( "For D in 2nd position, we have 3*6 possibilities
\n" ); document.write( "For D in 1st position we have 4*6\r
\n" ); document.write( "\n" ); document.write( "Total: 6 + 2*6 + 3*6 + 4*6 + 5*6 = 15 * 6 = 90\r
\n" ); document.write( "\n" ); document.write( "Combinatorics is not my specialty. You might want to check I did it right.
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