document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1002588>Question 1002588</A>: <I><SPAN STYLE=\"word-break: break-all;\"> please help to solve this question \r<BR>\n" );
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document.write( "in how many ways can 8 pupils sit in a row if two must not sit near each other </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #619460 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=619460\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the formula that i came up with is:\r<BR>\n" );
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document.write( "the number of ways to arrange where any 2 can't be together is:\r<BR>\n" );
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document.write( "n! - (n-1!) * 2\r<BR>\n" );
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document.write( "this is derived by figuring out the ways that have to be together and then subtracting that from the number of ways they can be arranged without restriction.\r<BR>\n" );
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document.write( "when they have to be together, they can be modelled as one unit.\r<BR>\n" );
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document.write( "when they are modelled as one unit, you get (n-1)! rather than n!, because there is one less member to consider since 2 of them are being treated as 1.\r<BR>\n" );
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document.write( "each unit pair, however, can be modelled in 2 ways, so you need to multiply that by 2.\r<BR>\n" );
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document.write( "you would get the number of ways any 2 have to be together would be (n-1)! * 2.\r<BR>\n" );
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document.write( "the number of ways that can't be together would then be n! - (n-1)! * 2.\r<BR>\n" );
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document.write( "that's the formula i came up with.\r<BR>\n" );
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document.write( "i modelled it with 3 members and with 4 members and it appears to be accurate.\r<BR>\n" );
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document.write( "i did not model it with 5 members bcause the possible number of combinations became too high to do manually.\r<BR>\n" );
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document.write( "i satisfied myself with the 3 member and 4 member models and figured the formula was probably good for more.\r<BR>\n" );
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document.write( "here's my work.\r<BR>\n" );
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document.write( "start with 3.\r<BR>\n" );
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document.write( "number of arrangements is 3! without any restrictions.\r<BR>\n" );
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document.write( "3! = 6\r<BR>\n" );
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document.write( "arrangements without any restrictions are:\r<BR>\n" );
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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\r<BR>\n" );
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document.write( "if 2 must be together then they act as 1 unit.<BR>\n" );
document.write( "you would then get 2! = 2<BR>\n" );
document.write( "since each unit has two possible arrangements, then multiply that by 2 to get 4.<BR>\n" );
document.write( "you have 4 possible arrangements out of 6 where they have to be together.<BR>\n" );
document.write( "6 - 4 = 2 possible arrangements where they can't be together.\r<BR>\n" );
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document.write( "marking the arrangements where a and b have to be together, you get:\r<BR>\n" );
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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 ***\r<BR>\n" );
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document.write( "there are 4 where they are together.<BR>\n" );
document.write( "what's left is 2 where they are not together.<BR>\n" );
document.write( "the formula is 3! - 2*2! = 6 - 4 = 2<BR>\n" );
document.write( "it works with 3 members.\r<BR>\n" );
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document.write( "now we'll do the same thing with 4 members.\r<BR>\n" );
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document.write( "number of arrangements without restrictions is 4! = 24\r<BR>\n" );
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document.write( "those arrangements are:\r<BR>\n" );
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document.write( "abcd<BR>\n" );
document.write( "abdc<BR>\n" );
document.write( "acbd<BR>\n" );
document.write( "acdb<BR>\n" );
document.write( "adbc<BR>\n" );
document.write( "adcb\r<BR>\n" );
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document.write( "bacd<BR>\n" );
document.write( "badc<BR>\n" );
document.write( "bcad<BR>\n" );
document.write( "bcda<BR>\n" );
document.write( "bdac<BR>\n" );
document.write( "bdca\r<BR>\n" );
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document.write( "cabd<BR>\n" );
document.write( "cadb<BR>\n" );
document.write( "cbad<BR>\n" );
document.write( "cbda<BR>\n" );
document.write( "cdab<BR>\n" );
document.write( "cdba\r<BR>\n" );
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document.write( "dabc<BR>\n" );
document.write( "dacb<BR>\n" );
document.write( "dbac<BR>\n" );
document.write( "dbca<BR>\n" );
document.write( "dcab<BR>\n" );
document.write( "dcba\r<BR>\n" );
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document.write( "we'll mark the arrangements where a and b have to be together.<BR>\n" );
document.write( "if the formula is correct, the number should be 3! * 2 = 6 * 2 = 12.\r<BR>\n" );
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document.write( "abcd ***<BR>\n" );
document.write( "abdc ***<BR>\n" );
document.write( "acbd<BR>\n" );
document.write( "acdb<BR>\n" );
document.write( "adbc<BR>\n" );
document.write( "adcb\r<BR>\n" );
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document.write( "bacd ***<BR>\n" );
document.write( "badc ***<BR>\n" );
document.write( "bcad<BR>\n" );
document.write( "bcda<BR>\n" );
document.write( "bdac<BR>\n" );
document.write( "bdca\r<BR>\n" );
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document.write( "cabd ***<BR>\n" );
document.write( "cadb<BR>\n" );
document.write( "cbad ***<BR>\n" );
document.write( "cbda<BR>\n" );
document.write( "cdab ***<BR>\n" );
document.write( "cdba ***\r<BR>\n" );
document.write( "<BR>\n" );
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document.write( "dabc ***<BR>\n" );
document.write( "dacb<BR>\n" );
document.write( "dbac ***<BR>\n" );
document.write( "dbca<BR>\n" );
document.write( "dcab ***<BR>\n" );
document.write( "dcba ***\r<BR>\n" );
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document.write( "the number where they have to be together is 12.<BR>\n" );
document.write( "24 - 12 = 12<BR>\n" );
document.write( "the number of ways where they must not be together is also 12.\r<BR>\n" );
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document.write( "the formula works with 3 and 4 members.<BR>\n" );
document.write( "it more then likely will work with 5 or more.<BR>\n" );
document.write( "i'll gamble and assume that it does.\r<BR>\n" );
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document.write( "the formula for numbr of arrangements where 2 members can't be together is:\r<BR>\n" );
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document.write( "n! - (n-1)! * 2\r<BR>\n" );
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document.write( "with 3 members, that becomes 3! - 2! * 2 = 6 - 2*2 = 6 - 4 = 2.\r<BR>\n" );
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document.write( "this was confirmed to be correct.\r<BR>\n" );
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document.write( "with 4 members, that becomes 4! - 3! * 2 = 24 - 6*2 = 24 - 12 = 12.\r<BR>\n" );
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document.write( "this was confirmed to be correct.\r<BR>\n" );
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document.write( "with 8 members, that becomes 8! - 7! * 2 = 40320 - 5040*2 = 40320 - 10080 = 30240.\r<BR>\n" );
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document.write( "this is assumed to be correct.<BR>\n" );
document.write( "there's no way of knowing for sure unless you can model it.<BR>\n" );
document.write( "you assumed it's good on faith that the formula applies to any number of members and not to just 3 and 4.\r<BR>\n" );
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document.write( "while it could be modelled manually, doing so would take way too long and the chance of error would be way too high.\r<BR>\n" );
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document.write( "some sort of software would have to be developed to do it mechanically.\r<BR>\n" );
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document.write( "i'll go with my gamble.<BR>\n" );
document.write( "hopefully the answer will be what you expect.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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