document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=583347>Question 583347</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many committees of two or more can be selected from 10 people? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #372493 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=372493\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The number of ways is 10C2 + 10C3 + ... + 10C10. If you know that 10C0 + 10C1 + 10C2 + ... + 10C10 = 2^10 = 1024, we can take this and subtract off 10C0 and 10C1, leaving 1024 - 1 - 10 = 1013. </SPAN>\n" );
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