document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1111291>Question 1111291</A>: <I><SPAN STYLE=\"word-break: break-all;\"> When testing for current in a cable with eight &#8203;color-coded wires, the author used a meter to test three wires at a time. How many different tests are required for every possible pairing of three &#8203;wires? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #726312 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=726312\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you would use the combination formula of:\r<BR>\n" );
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document.write( "c(n,x) = n! / (x! * (n-x)!)\r<BR>\n" );
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document.write( "n = 8<BR>\n" );
document.write( "x = 3\r<BR>\n" );
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document.write( "formula becomes c(8,3) = 8! / (3! * 5!)\r<BR>\n" );
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document.write( "result is c(8,3) = 56 different test would be required to get all possible combinations of 3 wires each, where order within each set of 3 is not important.\r<BR>\n" );
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document.write( "to see how this works, make n smaller so the number of possible combinations will be less.\r<BR>\n" );
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document.write( "when n = 4, c(4,3) = 4\r<BR>\n" );
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document.write( "let the wires be a, b, c, and d.\r<BR>\n" );
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document.write( "the test required to make sure you got all possible combinations would be:\r<BR>\n" );
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document.write( "abc<BR>\n" );
document.write( "abd<BR>\n" );
document.write( "acd<BR>\n" );
document.write( "bcd\r<BR>\n" );
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document.write( "note that the combination formula assumes order is not important within each set of 3.\r<BR>\n" );
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document.write( "if order is important, then the permutation formula would be used.\r<BR>\n" );
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document.write( "that formula is p(n,x) = n! / (n-x)!\r<BR>\n" );
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document.write( "note that the x! in the denominator is missing in the permutation formula, while it is present in the combination formula.\r<BR>\n" );
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document.write( "the x! in the denominator is the one that takes away the order is important part.<BR>\n" );
document.write( "that's what turns the permutation formula into the combination formula.\r<BR>\n" );
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document.write( "the number of possible permutations with p(4,3) is 4! / (4 - 3)! = 4! / 1! = 24\r<BR>\n" );
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document.write( "those permutations would be:\r<BR>\n" );
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document.write( "abc ***** original<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( "abd ***** original<BR>\n" );
document.write( "adb<BR>\n" );
document.write( "bad<BR>\n" );
document.write( "bda<BR>\n" );
document.write( "dab<BR>\n" );
document.write( "dba\r<BR>\n" );
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document.write( "acd ***** original<BR>\n" );
document.write( "adc<BR>\n" );
document.write( "cad<BR>\n" );
document.write( "cda<BR>\n" );
document.write( "dac<BR>\n" );
document.write( "dca\r<BR>\n" );
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document.write( "bcd ***** original<BR>\n" );
document.write( "bdc<BR>\n" );
document.write( "cbd<BR>\n" );
document.write( "cdb<BR>\n" );
document.write( "dbc<BR>\n" );
document.write( "dcb\r<BR>\n" );
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document.write( "out of the 24 possible permutations, there are only 4 possible combinations.\r<BR>\n" );
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document.write( "note that, for each of the originals that were derived from the combination formula, there are 6 possible permutations.\r<BR>\n" );
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document.write( "3! = 6, which is where the x! in the denominator comes from in the combination formula.\r<BR>\n" );
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document.write( "c(n,x) = n! / (x! * (n-x)!) becomes c(4,3) = 4! / (3! * (4-3)!) which becomes c(4,3) = 4! / (3! * 1!) = 4\r<BR>\n" );
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document.write( "p(n,x) = n! / (n-x)! becomes p(4,3) = 4! / (4-3)! which becomes p(4,3) = 4! / 1! = 24\r<BR>\n" );
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document.write( "bottom line:\r<BR>\n" );
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document.write( "you have 8 wires and you want to test all possible combinations of 3 of these at a time.\r<BR>\n" );
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document.write( "order, or arrangement of the wires within each set of 3 is not important, therefore you use the combination formula and not the permutations formula.\r<BR>\n" );
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document.write( "c(8,3) = 8! / (3! * 5!) = 56 possible sets of 3 wires where order within each set of 3 is not important.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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