document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=953596>Question 953596</A>: <I><SPAN STYLE=\"word-break: break-all;\"> in how many ways can a party of seven persons arrange themselves<BR>\n" );
document.write( "a. in a row of seven chairs?<BR>\n" );
document.write( "b. around a circular table?\r<BR>\n" );
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document.write( "using combinatorial analysis </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #582376 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=582376\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> In a row of 7 chairs, the number of permutations would be 7!.\r<BR>\n" );
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document.write( "around a circular table it becomes a little more complicated.\r<BR>\n" );
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document.write( "this is because there is no fixed first chair.\r<BR>\n" );
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document.write( "it could be any one of the 7.\r<BR>\n" );
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document.write( "i would estimate the number of permutations would be 7 * 7!.\r<BR>\n" );
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document.write( "but i could be wrong, so we'll have to do a demo to see what happens.\r<BR>\n" );
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document.write( "we can see how that works with less permutations to deal with.\r<BR>\n" );
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document.write( "7! = 5040 permutations.<BR>\n" );
document.write( "3! = 6 permutations.\r<BR>\n" );
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document.write( "3! is much easier to work with than 7!, so we'll use that as a demo.\r<BR>\n" );
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document.write( "let the seats be labeled a,b,c\r<BR>\n" );
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document.write( "in a straight line, the 6 possible permutations would be:\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( "now let's take a look at the circular table.\r<BR>\n" );
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document.write( "that first arrangement could be repeated 3 times in all.\r<BR>\n" );
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document.write( "first time starting with position 1.<BR>\n" );
document.write( "second time starting with position 2.<BR>\n" );
document.write( "third time starting with position 3.\r<BR>\n" );
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document.write( "you would get:\r<BR>\n" );
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document.write( "abc<BR>\n" );
document.write( "cab<BR>\n" );
document.write( "bca\r<BR>\n" );
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document.write( "a started in first position.<BR>\n" );
document.write( "a started in second position and then wrapped around.<BR>\n" );
document.write( "a started in third position and then wrapped around.\r<BR>\n" );
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document.write( "cab and bca are different arrangement.<BR>\n" );
document.write( "however, they are covered in the primary set that we got from the straight line arrangement, so there are no new arrangements.\r<BR>\n" );
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document.write( "let's take a look at the second arrangement of acb.\r<BR>\n" );
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document.write( "wrapping that arrangement around as we did the first, we would get:\r<BR>\n" );
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document.write( "acb<BR>\n" );
document.write( "bac<BR>\n" );
document.write( "cba\r<BR>\n" );
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document.write( "we have 2 new arrangements of bac and cba.\r<BR>\n" );
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document.write( "bac and cba are, however, already covered in the straight line permutations.\r<BR>\n" );
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document.write( "let's jump to the last arrangement of cba.\r<BR>\n" );
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document.write( "wrapping them around as we did the first of the circulars, we will get:\r<BR>\n" );
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document.write( "cba<BR>\n" );
document.write( "acb<BR>\n" );
document.write( "bca\r<BR>\n" );
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document.write( "we have 2 additional arrangement again.<BR>\n" );
document.write( "they are acb and bca.\r<BR>\n" );
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document.write( "they are also covered in the initial set of arrangements from the straight line version.\r<BR>\n" );
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document.write( "so it appears that the straight line arrangement and the circular arrangement give you the same number of permutations.\r<BR>\n" );
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document.write( "i made it even simpler to see  if we get the same result.\r<BR>\n" );
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document.write( "i assumed 2 arrangements called a and b.\r<BR>\n" );
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document.write( "straight line:\r<BR>\n" );
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document.write( "there are 2 possible permutations.\r<BR>\n" );
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document.write( "they are:\r<BR>\n" );
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document.write( "ab<BR>\n" );
document.write( "ba\r<BR>\n" );
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document.write( "i took ab and then rotated the positions around the table and got:\r<BR>\n" );
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document.write( "ab amd ba.\r<BR>\n" );
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document.write( "i took ba and then rotated the positions around the table and got:\r<BR>\n" );
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document.write( "ba and ab.\r<BR>\n" );
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document.write( "i got 4 arrangements but 2 of them were identical so that reduced to 2 distinct arrangements.\r<BR>\n" );
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document.write( "the number of permutations was the same whether i used a straight line presentation or a circular presentation.\r<BR>\n" );
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document.write( "i'll go with that.\r<BR>\n" );
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document.write( "you can do the same experiment for yourself to see if you get the same conclusion.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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