document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1028072>Question 1028072</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A veterinarian surveys 40 of his patrons. He discovers that 15 have dogs, 16 have cats, and 11 have fish. Six have dogs and cats, 8 have dogs and fish, and five have a cat and a fish. If two or them have all three kinds of pets, how many patrons have none of these pets? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #643261 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=643261\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You can solve this by using three intersecting circles...kinda hard to show here...but the advice is to start from the inside and work your way out...start from 2 in the central region where all three circles intersect and go from there...it isn't clear whether those that have dogs and cats can also have fish, but assuming that is so, I get 15 people have none of those... </SPAN>\n" );
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