document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=848561>Question 848561</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many 3 digit numbers can be formed from 1,2,3,4,5,6 if<BR>\n" );
document.write( "a) repition is allowed<BR>\n" );
document.write( "b) the number MUST be odd and no repetition is allowed<BR>\n" );
document.write( "c) the number MUST be less than 300 and no repition is allowed </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #511171 by </B><A HREF=/tutors/aboutme.mpl?userid=swincher4391><B>swincher4391(1107)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=swincher4391><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=swincher4391><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=511171\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> a) If repetition is allowed, this is a form of with replacement. We can calculate this pretty easily.   In short, there is 6^3 (216) ways to do this because for each number we place, there are 6 numbers that we can use.\r<BR>\n" );
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document.write( "b) Here this is considered without replacement. After the use of a digit, it no longer is a possible option for the next digit. So we have 6*5*4= 120 This commonly represented as 6P3.\r<BR>\n" );
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document.write( "c) Notice our first digit is now only limited to 1 and 2. So there is only 2 numbers that satisfy our first digit. Our second digit only has 5 possible digits. Our third digit only has 4 numbers that can satisfy it.\r<BR>\n" );
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document.write( "So 2*5*4 = 40 numbers </SPAN>\n" );
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