document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=116422>Question 116422</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Five girls- Fiorenza, Gladys, Helene, Jocelyn, and Kaitlin - and four boys- Abe, Bruce, Clive, and Doug ride to school each day in three seperate buses.<BR>\n" );
document.write( "Abe and Fiorenza always ride together.<BR>\n" );
document.write( "Gladys and Helen always ride together.<BR>\n" );
document.write( "Jocelyn and Kaitlin never ride together.<BR>\n" );
document.write( "Doug always rides in the bus with the fewest children.<BR>\n" );
document.write( "Boys cannot outnumber girls in any bus.<BR>\n" );
document.write( "The maximum number of children in any bus is four.\r<BR>\n" );
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document.write( "The bus in which Doug rides can hold how many children?\r<BR>\n" );
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document.write( "Bruce can ride with each of the following except<BR>\n" );
document.write( "a.Abe<BR>\n" );
document.write( "b.Clive<BR>\n" );
document.write( "c.Doug<BR>\n" );
document.write( "f.Fiorenza<BR>\n" );
document.write( "e.Helene </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #84648 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=84648\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> First thing we know is that there are three buses and nine children.  The situation where each bus carries three children is not possible because there are four boys, so one of the buses would have to carry two of the boys and on that bus the boys would outnumber the girls 2 to 1.\r<BR>\n" );
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document.write( "We are told that the most children on any bus is 4, so that gives us a high end to the range.  Could any of the buses carry just 1 passenger?  Since Doug rides on the bus with the fewest children, he would have to be that 1 passenger, but then he would outnumber the girls on that bus 1 to 0.  So a bus with only 1 passenger is not possible, and the bus with the fewest passengers, and the one that Doug rides, has 2 children.  The numbers of passengers have to be 4 for one bus, 3 for the second bus, and 2 for the remaining bus. (3, 3, and 3; 4, 4, and 1; 4, 3, and 2 are the only ways to divide 9 people into three groups where the largest group is 4)\r<BR>\n" );
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document.write( "Since Doug rides the bus with the fewest passengers, he must be on the bus with 2 passengers, and the \"Boys cannot outnumber girls in any bus\" rule means that the second passenger on this bus must be one of the girls.  That means that the second passenger can't be Bruce, and one answer to the second part of the question is c: Doug.\r<BR>\n" );
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document.write( "However, let's make certain that the other four responses to the second part of the problem are ok with the rules -- in other words, is it possible to create a combination of passengers including Bruce and the listed person that doesn't violate any of the conditions?\r<BR>\n" );
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document.write( "a: Abe.  Abe always rides with Fiorenza.  If Bruce were with Abe, then they couldn't be on the 3 person bus because it would be boys 2, girls 1.  So for Bruce and Abe to be together, they would have to be on the 4 person bus, and the fourth passenger would have to be a girl, perhaps Jocelyn or Kaitlin.  So this is possible. That would mean that Clive, Gladys, and Helene were on the 3 person bus, and Doug would be with whichever of Jocelyn or Kaitlin was not on the 4 person bus.\r<BR>\n" );
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document.write( "b: Clive:  Clive and Bruce together means that there would have to be two girls to ride with them and they would all be on the 4 person bus.  But this is possible because the two girls could be Gladys and Helene.  This would mean that Abe and Fiorenza were on the 3 person bus with either Jocelyn or Kaitlin, and Doug would be with Kaitlin or Jocelyn.\r<BR>\n" );
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document.write( "d: Fiorenza.  This is the same analysis as Bruce and Abe, therefore possible.\r<BR>\n" );
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document.write( "e: Helene:  Bruce could either be with Gladys and Helene on the 3 person bus, or with Gladys, Helene, and Clive on the 4 person bus.  Note that Clive is the only possibility for this configuration because if the fourth person were a girl, you couldn't keep the \"Boys cannot outnumber girls in any bus\" rule on the 3 person bus, Doug is on the 2 person bus already, and Abe has to be with Fiorenza.\r<BR>\n" );
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document.write( "Bruce, Gladys, and Helene on the 3 person bus means that Abe, Fiorenza, Clive, and either Jocelyn or Kaitlin were on the 4 person bus, and again, Doug is riding with either Kaitlin or Jocelyn\r<BR>\n" );
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document.write( "Bruce, Gladys, Helene, and Clive on the 4<BR>\n" );
document.write( "Abe, Fiorenza, and (Jocelyn or Kaitlin) on the 3<BR>\n" );
document.write( "Doug, (Kaitlin or Jocelyn) on the 2\r<BR>\n" );
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document.write( "So, the only correct answer to the second part of the question is c: Doug. </SPAN>\n" );
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