document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1120447>Question 1120447</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Sergio , Tyrone , Jim , and Dawn have all been invited to a dinner party. They arrive randomly and each person arrives at a different time.\r<BR>\n" );
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document.write( "a. In how many ways can they&#8203; arrive?<BR>\n" );
document.write( "b. In how many ways can Sergio arrive first and Dawn &#8203;last?<BR>\n" );
document.write( "c. Find the probability that Sergio will arrive first and Dawn last. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #736104 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13198)</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=736104\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "(a) 4! = 24 ways.  Any of the 4 first; then any of the remaining 3 next; then either of the other 2, then the last 1.<br><BR>\n" );
document.write( "(b) Either Tyrone or Jim can arrive second and then the other arrives third: 2 ways.<br><BR>\n" );
document.write( "(c) The answer to (b) divided by the answer to (a); 2 ways for that to happen, out of a total of 24 different possible ways: 2/24 = 1/12. </SPAN>\n" );
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