document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=879135>Question 879135</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many distinguishable ways can be written using all the letters in the word ALGEBRA?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #530531 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=530531\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Number of letters in this word are:<BR>\n" );
document.write( "A = 2<BR>\n" );
document.write( "L = 1<BR>\n" );
document.write( "G = 1<BR>\n" );
document.write( "E = 1<BR>\n" );
document.write( "B = 1<BR>\n" );
document.write( "R = 1<BR>\n" );
document.write( "There are 7 total letters with 2 of them being the same.<BR>\n" );
document.write( "The formula is 7! / 2! = 2520.\r<BR>\n" );
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document.write( "This is hard to show because there are so many possibilities, but I can show it with a simpler type problem.\r<BR>\n" );
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document.write( "Assume the letters ABC.<BR>\n" );
document.write( "The  number of possible arrangements are 3! = 6<BR>\n" );
document.write( "Those arrangements 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\r<BR>\n" );
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document.write( "Now assume the letters AAC.<BR>\n" );
document.write( "The number of possible arrangements are 3! / 2! = 3.<BR>\n" );
document.write( "Those arrangements are:<BR>\n" );
document.write( "AAC<BR>\n" );
document.write( "ACA<BR>\n" );
document.write( "CAA\r<BR>\n" );
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document.write( "You have essentially replace the B with another A.<BR>\n" );
document.write( "Where you had ABC and ACB, you now only have AAC and ACA<BR>\n" );
document.write( "Where you had BAC and BCA, you now only have AAC and ACA<BR>\n" );
document.write( "Where you had CAB and CBA, you now only have CAA and CAA.<BR>\n" );
document.write( "You now have AAC twice and ACA twice and CAA twice.<BR>\n" );
document.write( "The number of unique arrangements is only 3 which are AAC and ACA and CAA.\r<BR>\n" );
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document.write( "The general formula for number of unique permutations is:\r<BR>\n" );
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document.write( "P = n! / (x1! * x2! * ... xn!) where:\r<BR>\n" );
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document.write( "n is the number of letters.<BR>\n" );
document.write( "x1, x2, ..., xn are the number of letters that are the same.\r<BR>\n" );
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document.write( "For example:\r<BR>\n" );
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document.write( "Original letters are ABCDEFG\r<BR>\n" );
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document.write( "Formula is 7!\r<BR>\n" );
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document.write( "Original letters are AACCEEG\r<BR>\n" );
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document.write( "Formula is 7! / (2! * 2! * 2!)\r<BR>\n" );
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document.write( "Original letters are AAAAABB\r<BR>\n" );
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document.write( "Formula is 7! / (5! * 2!)\r<BR>\n" );
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document.write( "That's how it works.\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );