document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1102309>Question 1102309</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 2 math books,4 science books and 3 literature books are to be arranged on a shelf. In how many ways can this be done if the 2 math books are not together? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #717189 by </B><A HREF=/tutors/aboutme.mpl?userid=math_helper><B>math_helper(2461)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_helper><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=math_helper><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=717189\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Ignoring the math book restriction, there are  9! = 362880 unique ways of arranging the books.<BR>\n" );
document.write( "Now we need to divide that by the number of arrangements where  the two math books are together:<BR>\n" );
document.write( "There are 8 ways for m1m2 to appear and 8 ways for m2m1 to appear, so we need to divide by 16:\r<BR>\n" );
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document.write( "362880/16 = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+highlight%2822680%29+ ALIGN=MIDDLE  ALT=\"+highlight%2822680%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "—<BR>\n" );
document.write( "EDIT 11/25:  Yes, my answer over-deducts (i.e. dividing by 16 takes out too many arrangements) for the M1M2, M2M1 cases.   9! - 2*8!  removes the arrangements properly for M1M2 & M2M1 together.    [ If I had sanity-checked my answer I'd have seen that it is way too small. ] <BR>\n" );
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