document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=463077>Question 463077</A>: <I><SPAN STYLE=\"word-break: break-all;\"> pls help... answer ASAP :   \r<BR>\n" );
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document.write( "a three digit code is made up of three diff. digts from the set (2,4,6,8,) . how many three digit codes can be formed if no digit can be repeated?... pls explain and show the solution...... </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #317332 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=317332\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the set of digits is 2,4,6,8\r<BR>\n" );
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document.write( "you are taking a set of 3 out of a set of 4.\r<BR>\n" );
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document.write( "if order is not important, then you are taking a combination of 3 out of 4.\r<BR>\n" );
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document.write( "the formula for that is n! / (x! * (n-x)!)( which equals 4! / (1! * 3!) which becomes 4.\r<BR>\n" );
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document.write( "if order is important, than you are taking a permutation of 3 out of 4.\r<BR>\n" );
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document.write( "the formula for that is n! / (n-x)! which equals 4! / 1! which equals 24.\r<BR>\n" );
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document.write( "the number of combinations is the number of sets that have at least one element different from any other set.\r<BR>\n" );
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document.write( "your possible combinations are:\r<BR>\n" );
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document.write( "246<BR>\n" );
document.write( "248<BR>\n" );
document.write( "268<BR>\n" );
document.write( "468\r<BR>\n" );
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document.write( "within each of these sets, you can make 6 arrangements by changing the order of each element in the set.\r<BR>\n" );
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document.write( "example:\r<BR>\n" );
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document.write( "246 possible arrangements are:\r<BR>\n" );
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document.write( "246<BR>\n" );
document.write( "264<BR>\n" );
document.write( "426<BR>\n" );
document.write( "462<BR>\n" );
document.write( "624<BR>\n" );
document.write( "642\r<BR>\n" );
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document.write( "similarly you can do the same with with the other 3 sets.\r<BR>\n" );
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document.write( "4 sets with 6 arrangements within each set gets you the 24 permutations.\r<BR>\n" );
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document.write( "if the digits could be repeated, the formula would be changed as follows:\r<BR>\n" );
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document.write( "you would have 4^3 possible arrangments which would be equal to 64 possible arrangements.\r<BR>\n" );
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document.write( "the first number could be any of 2,4,6,8<BR>\n" );
document.write( "the second number could be any of 2,4,6,8<BR>\n" );
document.write( "the third number could be any of 2,4,6,8\r<BR>\n" );
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document.write( "4 possible arrangments for each digit gets you the 4^3.\r<BR>\n" );
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document.write( "examples would be:\r<BR>\n" );
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document.write( "222<BR>\n" );
document.write( "223<BR>\n" );
document.write( "233<BR>\n" );
document.write( "333<BR>\n" );
document.write( "334<BR>\n" );
document.write( "344<BR>\n" );
document.write( "444<BR>\n" );
document.write( "445<BR>\n" );
document.write( "455<BR>\n" );
document.write( "555<BR>\n" );
document.write( "etc,\r<BR>\n" );
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document.write( "showing you 64 possible arrangements is tedious.\r<BR>\n" );
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document.write( "assume you had 3 numbers and you chose 2 out of 3.\r<BR>\n" );
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document.write( "if the numbers could be repeated than you could have 3^2 = 9 possible arrangements\r<BR>\n" );
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document.write( "let the numbers be 1, 2, 3, and the possible arrangements would be:\r<BR>\n" );
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document.write( "11<BR>\n" );
document.write( "12<BR>\n" );
document.write( "13<BR>\n" );
document.write( "21<BR>\n" );
document.write( "22<BR>\n" );
document.write( "23<BR>\n" );
document.write( "31<BR>\n" );
document.write( "32<BR>\n" );
document.write( "33\r<BR>\n" );
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document.write( "this assumes the same numbers can be used more than once.\r<BR>\n" );
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document.write( "if the same numbers could not be used more than once, then the formulas shown above would apply.\r<BR>\n" );
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document.write( "the number of combinations would be n! / (x! * (n-x)!) which would be equal to 3! / (2! * 1!) which would be equal to 6 / 2 which would be equal to 3.\r<BR>\n" );
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document.write( "those combinations would be:\r<BR>\n" );
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document.write( "12<BR>\n" );
document.write( "13<BR>\n" );
document.write( "23\r<BR>\n" );
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document.write( "the number of permutations would be n! / (n-x)! which would be equal to 3! / 1! which would be equal to 6.\r<BR>\n" );
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document.write( "those permutations would be:\r<BR>\n" );
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document.write( "12<BR>\n" );
document.write( "21<BR>\n" );
document.write( "13<BR>\n" );
document.write( "31<BR>\n" );
document.write( "23<BR>\n" );
document.write( "31\r<BR>\n" );
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document.write( "because the numbers were smaller, i could show you the details of how this works out.\r<BR>\n" );
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document.write( "in your problem, the answer is:\r<BR>\n" );
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document.write( "number of combinations equals 4! / (3! * 1!) = 4\r<BR>\n" );
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document.write( "number of permutations equals 4! / (1!) = 4 * 3 * 2 = 24\r<BR>\n" );
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document.write( " </SPAN>\n" );
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