document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=602213>Question 602213</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A committee of 5 is selected from 5 men and 6 women.  How many committees are possible if there must be at least 3 men of the committee?\r<BR>\n" );
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document.write( "[181 is the correct answer, however, I'm not able to get that answer by multi. nCr of my groups of (5,5) and (6,3).  Please explain how to get this answer.   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #380158 by </B><A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><B>jim_thompson5910(35256)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jim_thompson5910><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=380158\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> There are (5 C 3)*(6 C 2) = 10*15 = 150 ways to have exactly 3 men on the committee.\r<BR>\n" );
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document.write( "There are (5 C 4)*(6 C 1) = 5*6 = 30 ways to have exactly 4 men on the committee.\r<BR>\n" );
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document.write( "and\r<BR>\n" );
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document.write( "There is (5 C 5)*(6 C 0) = 1*1 = 1 way to have exactly 5 men on the committee.\r<BR>\n" );
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document.write( "Add these results up: 150+30+1 = 181\r<BR>\n" );
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document.write( "So there are <font color=\"red\">181</font> ways to form the committee if there must be at least 3 men on the committee. </SPAN>\n" );
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