document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=48934>Question 48934</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how many numbers of five different digits, each number to contain 3 odd and 2 even digits, can be formed from the digits 1,2,3,4,5,6,7,8 and 9?Thanks! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #32409 by </B><A HREF=/tutors/aboutme.mpl?userid=longjonsilver><B>longjonsilver(2297)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=longjonsilver><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=longjonsilver><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=32409\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> you need to build up the solution.\r<BR>\n" );
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document.write( "Even numbers: 2,4,6,8<BR>\n" );
document.write( "Odd numbers: 1,3,5,7,9\r<BR>\n" );
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document.write( "The 5 digit number: XXXYY has to have 3 odds (the X's). So, lets look at just those:\r<BR>\n" );
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document.write( "Pick the first odd number. How many possibles are there? Answer 5 (1,3,5,7,9)<BR>\n" );
document.write( "Now pick the second odd number. How many possibles are there now? Answer 4 (since one of them was picked to be the first number.<BR>\n" );
document.write( "Similarly, the third odd number... there are 3 possibles.\r<BR>\n" );
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document.write( "So altogether, there are 5*4*3 --> 60 3-digit variations from 5 odd numbers.\r<BR>\n" );
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document.write( "Now for the evens. For a 2-digit number, the first of the 2 even numbers could be 1 of 4 numbers: answer = 4<BR>\n" );
document.write( "The final even number must be one of the remaining 3 evens: answer = 3\r<BR>\n" );
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document.write( "So, altogether there are 4*3 --> 12 2-digit variations from 4 even numbers.\r<BR>\n" );
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document.write( "And now we combine them... we know how many odd permutations there are (60) and even permutations (12). Combined, there are 60*12 --> 720 permutations of 3 odd and 2 even.\r<BR>\n" );
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document.write( "As you can see... at no point did i use any \"formula\". I just worked it out from \"first principles\" and \"common sense\"\r<BR>\n" );
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document.write( "Hope this helps.<BR>\n" );
document.write( "jon.<BR>\n" );
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