Question 921215
<pre>
We start with this Venn diagram.  The little letters from d through k
represent the number of elements in each of the 8 regions.  We are to 
fill in numbers for each of the question marks:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=?"),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,C),
locate(-.45,-1,"j=?"),
locate(.6,.4,"i=?"), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
locate(-1.5,.5,"g=?"),
locate(-.4,2.3,"e=?"),
locate(1.8,2,"f=?"),
locate(-.4,1.1,"h=?") )}}}

I notice that they are listed in the exact reverse order from the 
order that we will need them.  It's common for teachers and books to
list them backwards.

We begin with last one given. 

n(A &#8745; B &#8745; C) = 4

A &#8745; B &#8745; C is the set of elements that are COMMON to all three circles.
That is the region right in the center with h=? in it.  So h=4. We
write that in the middle:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=?"),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,C),
locate(-.45,-1,"j=?"),
locate(.6,.4,"i=?"), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
locate(-1.5,.5,"g=?"),
locate(-.4,2.3,"e=?"),
locate(1.8,2,"f=?"),
red(locate(-.4,1.1,"h=4")) )}}}

We now look at n(B &#8745; C) = 15.
B &#8745; C consists of the regions common to circles B and C.
That's the regions h and i. So h+i=15, and since h=4,
4+i=15 or i=11.  So we write i=11:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=?"),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,C),
locate(-.45,-1,"j=?"),
red(locate(.6,.4,"i=11")), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
locate(-1.5,.5,"g=?"),
locate(-.4,2.3,"e=?"),
locate(1.8,2,"f=?"),
red(locate(-.4,1.1,"h=4")) )}}}

We now look at n(A &#8745; C) = 12.
A &#8745; C consists of the regions common to circles A and C.
That's the regions g and h. So g+h=12, and since h=4,
g+4=12 or g=8.  So we write g=8:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=?"),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,C),
locate(-.45,-1,"j=?"),
red(locate(.6,.4,"i=11")), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
red(locate(-1.5,.5,"g=8")),
locate(-.4,2.3,"e=?"),
locate(1.8,2,"f=?"),
red(locate(-.4,1.1,"h=4")) )}}}

We now look at n(A &#8745; B) = 7.
A &#8745; B consists of the regions common to circles A and B.
That's the regions e and h. So e+h=7, and since h=4,
e+4=7 or e=3.  So we write e=3:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=?"),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,C),
locate(-.45,-1,"j=?"),
red(locate(.6,.4,"i=11")), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
red(locate(-1.5,.5,"g=8")),
red(locate(-.4,2.3,"e=3")),
locate(1.8,2,"f=?"),
red(locate(-.4,1.1,"h=4")) )}}}

We now look at n(C) = 31.
C consists of all the regions that make up circle C.
That's the four regions g, h, i and j. So g+h+i+j=31, and since g=8,
h=4, and i=11, then 8+4+11+j=31, 23+j=31  So j=8 and we write j=8:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=?"),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,C),
red(locate(-.45,-1,"j=8")),
red(locate(.6,.4,"i=11")), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
red(locate(-1.5,.5,"g=8")),
red(locate(-.4,2.3,"e=3")),
locate(1.8,2,"f=?"),
red(locate(-.4,1.1,"h=4")) )}}}

We now look at n(B) = 27.
B consists of all the regions that make up circle B.
That's the four regions e, f, h and i. So e+f+h+i=27, and since e=3,
h=4, and i=11, then 3+f+4+11=27, 18+f=27  So f=9 and we write f=9:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
locate(-2.3,2,"d=?"),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,C),
red(locate(-.45,-1,"j=8")),
red(locate(.6,.4,"i=11")), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
red(locate(-1.5,.5,"g=8")),
red(locate(-.4,2.3,"e=3")),
red(locate(1.8,2,"f=9")),
red(locate(-.4,1.1,"h=4")) )}}}

We now look at n(A) = 25.
B consists of all the regions that make up circle A.
That's the four regions d, e, g and h. So d+e+g+h=25, and since e=3,
g=8, and h=4, then d+3+8+4=25, d+15=25  So d=10 and we write d=10:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
red(locate(-2.3,2,"d=10")),
locate(-3.5,-2,"k=?"),
locate(0,-2.7,C),
red(locate(-.45,-1,"j=8")),
red(locate(.6,.4,"i=11")), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
red(locate(-1.5,.5,"g=8")),
red(locate(-.4,2.3,"e=3")),
red(locate(1.8,2,"f=9")),
red(locate(-.4,1.1,"h=4")) )}}}

Finally we look at n(U) = 150.  That's the whole set enclosed
in the large rectangle.  It consiste of all the regions,
so d+e+f+g+h+i+j+k = 150.  Substituting for all the letters 
except k, 10+3+9++8+4+11+8+k = 150.
                        53+k = 150
                           k = 97

So we fill in 97 and we have the entire Venn diagram filled in:

{{{drawing(300,300,-4,4,-5,4,
rectangle(-4,-3.5,4,4),
circle(0,-.5,2),
red(locate(-2.3,2,"d=10")),
red(locate(-3.5,-2,"k=97")),
locate(0,-2.7,C),
red(locate(-.45,-1,"j=8")),
red(locate(.6,.4,"i=11")), 
circle(sqrt(2),sqrt(2),2), 
locate(-3.5,2.5,A),
circle(-sqrt(2),sqrt(2),2),
locate(3.5,2.5,B),
red(locate(-1.5,.5,"g=8")),
red(locate(-.4,2.3,"e=3")),
red(locate(1.8,2,"f=9")),
red(locate(-.4,1.1,"h=4")) )}}}

Now we are ready to answer anything about the Venn diagram:

Since A consists of d,e,g,h, I will use the notation A = degh.
Similarly for the others:

n[A &#8745; (B &#8746; C)]
n[degh &#8745; (efhi &#8746; ghij)]
In the ( ) we have the union, which includes everything on 
either side of the &#8746;, or efghij.  So we have:

n[degh &#8745; (efghij)]

In the [ ] we have the intersection, which includes ONLY what 
is IN COMMON to the sets on the two sides of the &#8745;, or egh.
So we have:

n[egh] = 3+8+4 = 15  

--------------------

n[A &#8745; (B &#8746; C)<sup>c</sup>]

n[degh &#8745; (efhi &#8746; ghij)<sup>c</sup>]

n[degh &#8745; (efghij)<sup>c</sup>]

The superscript "c" tells us that we are
not to use any regions indicated in the 
parentheses, but are to use all the other
regions in the universal set instead. So
we must substitute dk for (efghij)<sup>c</sup>
since d and k are the only ones besides efghij.
So we have

n[degh &#8745; dk]

In the [ ] we have the intersection, which includes ONLY what 
is IN COMMON to the sets on the two sides of the &#8745;, or just d.
So we have:

n[d] = 10

Edwin</pre>