document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=475863>Question 475863</A>: <I><SPAN STYLE=\"word-break: break-all;\"> There are 10 points in a plane.  No three of these points are in a straight line, except 4 points which are all in the same straight line.  How many straight lines can be formed by joining the 10 points? \r<BR>\n" );
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document.write( "Here is how I did it: <BR>\n" );
document.write( "n=10 = points total<BR>\n" );
document.write( "r=2 = points required to make a line<BR>\n" );
document.write( "10C2 = 45 lines\r<BR>\n" );
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document.write( "n=4 = colinear points<BR>\n" );
document.write( "r=2 = points required to make a line<BR>\n" );
document.write( "4C2= 6 lines made from colinear points\r<BR>\n" );
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document.write( "n=6 = points that are not colinear<BR>\n" );
document.write( "r=2 = points required to make a line<BR>\n" );
document.write( "6C2 = 16 lines made from points which are not colinear\r<BR>\n" );
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document.write( "But I am not sure what to do with this now. If I multiply 4C2 with 6C2 I get 15*6 = 90 lines. It does not make sense to multiply.  Let's do 15 plus 6 that gives 21 lines are made from 10 points but that also sounds wrong; it's too small.  \r<BR>\n" );
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document.write( "Can I approach this problem this way?<BR>\n" );
document.write( "case 1<BR>\n" );
document.write( "n=4 colinear points<BR>\n" );
document.write( "each colinear point can make 6 lines; thus 1 linear point/6 lines, thus 4*6 = 24 lines\r<BR>\n" );
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document.write( "case 2<BR>\n" );
document.write( "n=6 (other points, not alligned)<BR>\n" );
document.write( "each not alligned point can make 9 lines, thus 1 non-alligned point/9 lines, thus 6*9 = 54 lines\r<BR>\n" );
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document.write( "case 1 + case 2 gives 54+24 = 78 lines can be formed by joining 4 colinear points and 6 not alligned points. But this is not the right answer either. \r<BR>\n" );
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document.write( "Would you explain please how to proceed with this problem?  Thank you so much for your help. \r<BR>\n" );
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document.write( "Yours, \r<BR>\n" );
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document.write( "I.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #326393 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=326393\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I think the simplest way is to start with the four collinear points, in which we have one line.\r<BR>\n" );
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document.write( "The next case would be involving the other six points: choose two of these six points, 6C2 = 15 ways to do it.\r<BR>\n" );
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document.write( "The final case would be to choose one of the four collinear points, then one of the six remaining points, in which 24 lines are determined (we do not overcount lines because no three of the other points are collinear).\r<BR>\n" );
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document.write( "The number of lines is 1+15+24 = 40. </SPAN>\n" );
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