document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=594320>Question 594320</A>: <I><SPAN STYLE=\"word-break: break-all;\"> simplify  (n-1)C(r-1) + (n-1)Cr </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #376863 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=376863\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Pascal's triangle. It can be proven using a committee-forming argument (suppose you choose r people from a group of n. If person X is chosen, we have to choose r-1 people from n-1. If X is not chosen, we choose r people from n-1).\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE {{n-1} \choose {r-1}} + {{n-1} \choose {r}} = {{n} \choose {r}}\">\r<BR>\n" );
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document.write( "You can also do it algebraically. </SPAN>\n" );
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