document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=366816>Question 366816</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 4)Eight policemen are to be posted to guard three separate buildings. <BR>\n" );
document.write( "In how many ways may they all be posted if no building is to be guarded by less than two policemen? \r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #261662 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=261662\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The partitioning can only take place in two ways: either 2,2,4 or 3,3,2.<BR>\n" );
document.write( "For the first case, there are C(8,4)*C(4,2)*C(2,2) = 420 ways of partitioning 8 men in two groups of two and 1 group of 4.  After partitioning the 3 groups can be assigned in 3! = 6 ways.  Therefore there are 420*6 = 2,520 ways.<BR>\n" );
document.write( "Similarly  there are C(8,3)*C(5,3)*C(2,2) = 560 ways of partitioning 8 men in two groups of 3 and 1 group of 2, and 6 ways of assigning them to the different buildings. Therefore there are 560*6 = 3,360 ways.<BR>\n" );
document.write( "Finally since the two cases are mutually exclusive, there are a total of 2,520 + 3,360 = 5,880 ways. </SPAN>\n" );
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