document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1129131>Question 1129131</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Three cards labelled 2, 3 and 4 are placed in a hat and a card is drawn. This is repeated 3 times without replacement. <BR>\n" );
document.write( "a) How many 3 digit numbers can be formed<BR>\n" );
document.write( "b) What is the probability that the 3 digit number is odd? divisible by 3?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #745698 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=745698\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Since the cards are drawn without replacement, there are 3 possibilities for <BR>\n" );
document.write( "the 1st number, 2 possibilities for the 2nd number and 1 possibility for the <BR>\n" );
document.write( "remaining number.<BR>\n" );
document.write( "Thus, there are 3x2x1 = 6 combinations<BR>\n" );
document.write( "Since the digits 2, 3 and 4 add up to 9, all the numbers are divisible by 9 <BR>\n" );
document.write( "and therefore also divisible by 3.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );