document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1093789>Question 1093789</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The total no. Of no. Less than 1000 and divisible by 5 formed with 0,1,2.......9 such that each digit does not occur more than once in each no.is  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #708412 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=708412\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br>If the number is divisible by 5, then the last digit is either 0 or 5.  So you have two cases to analyze.<br><BR>\n" );
document.write( "If the last digit is 0, then there are 9 choices for one of the other digits and then 8 choices for the third digit.  By the fundamental counting principle, the number of numbers less than 1000, divisible by 5, ending in 0, and containing 3 different digits is 9*8 = 72.<br><BR>\n" );
document.write( "If the last digit is 5, there are still 9 choices for one of the other digits.  That is because the problem said \"numbers less than 1000\", not \"3-digit numbers\" -- so the first digit can be 0.  And then again there are 8 choices for the last digit, making again 9*8=72 numbers less than 1000, divisible by 5, ending in 5, and containing 3 different digits.<br><BR>\n" );
document.write( "So there are a total of 72+72=144 numbers that are less than 1000, are divisible by 5, and contain 3 different digits. </SPAN>\n" );
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