document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=4083>Question 4083</A>: <I><SPAN STYLE=\"word-break: break-all;\">   In how many ways can a committee of 3 be selected from a group of 5?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #1826 by </B><A HREF=/tutors/aboutme.mpl?userid=rapaljer><B>rapaljer(4671)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rapaljer><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=rapaljer><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=1826\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is a combination problem, since the order of selection is not significant.  Standard notation for this can be written C(5,3), and most standard calculators can be used to do the calculation without even understanding it.  \r<BR>\n" );
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document.write( "However, understanding it, you can calculate the product of three numbers beginning at 5 and counting down:  5 times 4 times 3, and then divide by the product of the three numbers beginning at 3 and counting down:  3 times 2 times 1.  \r<BR>\n" );
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document.write( "Final answer is (5*4*3) over (3*2*1) or 60/6 = 10. </SPAN>\n" );
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