document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1182644>Question 1182644</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Consider the 720 permutations of the numbers 1, 2, 3, 4, 5, 6. In how many of<BR>\n" );
document.write( "these permutations does 1 appear next to 2 or 3 appear next to 4 or 5 appear<BR>\n" );
document.write( "next to 6? </SPAN>\n" );
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document.write( "Let A be the event that 1 and 2 appear next to each other, B be the event that 3 and 4 appear next to each other, and C be the event that 5 and 6 appear next to each other.<br><BR>\n" );
document.write( "Then use the inclusion-exclusion principle:<br><BR>\n" );
document.write( "# ways (A or B or C)<BR>\n" );
document.write( "= ((# of ways A) + (# of ways B) + (# of ways C))<BR>\n" );
document.write( "- ((# of ways A and B) + (# of ways A and C) + (# of ways B and C))<BR>\n" );
document.write( "+ (# of ways A and B and C)<br><BR>\n" );
document.write( "(1) Number of ways A can happen:<BR>\n" );
document.write( "Treat the 1 and 2 as a unit; then there are 5 items to be arranged -- the 1-2 pair and the other 4 digits.  The 5 items can be arranged in any of 5!=120 ways; the two items in the 1-2 pair can be arranged in either of 2!=2 ways.  The number of ways A can happen is 120*2=240.<br><BR>\n" );
document.write( "Then by the same analysis the number of ways B can happen and the number of ways C can happen are each 240.<br><BR>\n" );
document.write( "(2) Number of ways A and B can both happen:<BR>\n" );
document.write( "Treat the 1 and 2 as a unit and the 3 and 4 as a unit.  Then we have 4 items to arrange -- the two pairs and the other two digits.  The 4 items can be arranged in 4!=24 ways; each of the pairs can be arranged in either of 2!=2 ways.  The number of ways A and B can happen is 24*2*2 = 96.<br><BR>\n" );
document.write( "Then by the same analysis the number of ways A and C can happen is 96, and the number of ways B and C can happen is 96.<br><BR>\n" );
document.write( "(3) Number of ways A and B and C can all happen:<BR>\n" );
document.write( "Now we have three groups of 2 digits each.  Those three items can be arranged in 3!=6 ways; the two digits in each pair can be arranged in either or 2!=2 ways.  The number of ways A and B and C can all happen is 6*2*2*2=48.<br><BR>\n" );
document.write( "ANSWER: The number of ways 1 and 2 can be together or 3 and 4 can be together or 5 and 6 can be together is<br><BR>\n" );
document.write( "3(240)-3(96)+48 = 720-288+48 = 480<br><BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );