document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=760213>Question 760213</A>: <I><SPAN STYLE=\"word-break: break-all;\"> On a shelf there are 4 different mathematics and 3 different additional maths books. In how many ways can this be done if the additional maths books are together. I have tried everything but in vain, the answer should be 724.<BR>\n" );
document.write( "here's my workings [M][M][M][M] we can put the three additional maths books between them, and then I get this 3!*5!=720 but the answer is 724... </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #462481 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=462481\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> if the math books are together then they can be arranged in 3! ways = 6.<BR>\n" );
document.write( "if you put them on the shelf and they have to be together, then they act as if they're 1 so you would get 5! = 120.<BR>\n" );
document.write( "6 * 120 = 720<BR>\n" );
document.write( "i'm inclined to think that you're correct and the book is wrong.<BR>\n" );
document.write( "i tried this with 4 books where 2 of them have to be together.<BR>\n" );
document.write( "I labeled the books ABC where C stood for the pair of books that had to be together.<BR>\n" );
document.write( "I labeled the 2 books that had to be together as 1, 2.<BR>\n" );
document.write( "this is what I got:<BR>\n" );
document.write( "With ABC, you get 6 possible permutations.<BR>\n" );
document.write( "they are:<BR>\n" );
document.write( "ABC<BR>\n" );
document.write( "ACB<BR>\n" );
document.write( "BAC<BR>\n" );
document.write( "BCA<BR>\n" );
document.write( "CAB<BR>\n" );
document.write( "CBA<BR>\n" );
document.write( "when you replace C with 12, you get:<BR>\n" );
document.write( "AB12<BR>\n" );
document.write( "A12B<BR>\n" );
document.write( "BA12<BR>\n" );
document.write( "B12A<BR>\n" );
document.write( "12AB<BR>\n" );
document.write( "12BA<BR>\n" );
document.write( "when you permute 12, you get 1,2 and 2,1.<BR>\n" );
document.write( "the other half of the permutations would be:<BR>\n" );
document.write( "AB21<BR>\n" );
document.write( "A21B<BR>\n" );
document.write( "BA21<BR>\n" );
document.write( "B21A<BR>\n" );
document.write( "21AB<BR>\n" );
document.write( "21BA<BR>\n" );
document.write( "that's a total of 12 permutations which is equal to 3! * 2!.<BR>\n" );
document.write( "i don't see any other possible permutation.<BR>\n" );
document.write( "for AB, you can have AB12, AB21, A12B, A21B, 12AB, 21AB.<BR>\n" );
document.write( "for BA, you can have BA12, BA21, B12A, B21A, 12BA, 21BA.<BR>\n" );
document.write( "the books 12 have to be together but the books AB do not.<BR>\n" );
document.write( "if it works for 2 and 2, then it should work for 4 and 3.<BR>\n" );
document.write( "5! * 3! looks to me like the correct answer.<BR>\n" );
document.write( "i have no idea where the additional 4 would come from.<BR>\n" );
document.write( "i can't see it in the simple example.<BR>\n" );
document.write( "it's unlikely they exist in the more complex example.<BR>\n" );
document.write( "i'd stick with 720 even if the book says 724.<BR>\n" );
document.write( "the books are not always right.<BR>\n" );
document.write( "some books are more right than others.\r<BR>\n" );
document.write( "\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );