document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1086412>Question 1086412</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A committee of 6 persons to be formed from 4 men and 6 women.In how many ways can this be done when atmost 3 women are included? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #700604 by </B><A HREF=/tutors/aboutme.mpl?userid=math_helper><B>math_helper(2461)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_helper><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=math_helper><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=700604\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <BR>\n" );
document.write( "1.  If the committee has exactly 3 women:\r<BR>\n" );
document.write( "\n" );
document.write( "#ways of selecting 3 women * # ways of selecting 6-3=3 men:<BR>\n" );
document.write( "(6C3) x (4C3) = 6!/(3!*3!) * 4!/(3!*1!) = 20 * 4 = 80  ways \r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "2.  If the committee has exactly 2 women:<BR>\n" );
document.write( "(6C2) x (4C4) = 6!/(4!*2!) * 4!/(4!0!) = 15 * 1 = 15  ways\r<BR>\n" );
document.write( "\n" );
document.write( "Those are the only two possibilities for #women and #men because if you tried to form the committee with just 1 woman, there wouldn't be enough men to make a committee of 6.\r<BR>\n" );
document.write( "\n" );
document.write( "Thus, the answer is  80+15 =  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+highlight%2895%29+ ALIGN=MIDDLE  ALT=\"+highlight%2895%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > ways </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );