Question 459075
<pre><font face = "consolas" size = 2 color = "indigo"><b>
 
We will draw a Venn diagram consisting of 3 overlapping
circles A, B, C and let the letters t,u,v,w,x,y,z indicate the
number of elements that are in each of the 7 regions created
by the overlapping circles.

{{{drawing(300,300,-4,4,-5,4,
circle(0,-.5,2),
locate(-2,2,t),
locate(0,-2.7,C),
locate(-.3,-1,z),
locate(1.1,.4,y), 
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(-1.3,.5,w),
locate(0,2.5,u),
locate(2,2,v),
locate(-.2,1.1,x) )}}}

n(A) = t+u+w+x
n(B) = u+v+x+y
n(C) = w+x+y+z
n(A&#5198;B) = u+x = 6
n(A&#5198;C) = w+x = 8
n(B&#5198;C) = x+y = 7
n(A&#5198;B&#5198;C) = x = 4
n(A&#5196;B) = t+u+v+w+x+y
n(A&#5196;C) = t+u+w+x+y+z
n(B&#5196;C) = u+v+w+x+y+z = 20
n(A&#5196;B&#5196;C) = t+u+v+w+x+y+z = 25
n(B-A) = v+y = 5 
 
 
Only one of those given pieces of data
consists of only one of the 7 regions.
That one is 

n(A&#5198;B&#5198;C)= x = 4

so we will replace x by 4 in the region right in the
middle.

{{{drawing(300,300,-4,4,-5,4,

circle(0,-.5,2),
locate(-2,2,t),
locate(0,-2.7,C),
locate(-.3,-1,z),
locate(1.1,.4,y), 
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(-1.3,.5,w),
locate(0,2.5,u),
locate(2,2,v),
locate(-.2,1.1,4) )}}}

n(A&#5198;B) = u+x = 6
n(A&#5198;C) = w+x = 8
n(B&#5198;C) = x+y = 7

Since we know that n(A&#5198;B) = u+x = u+4 = 6
we know that u=2 so we replace u by 2

Also since we know that n(A&#5198;C) = w+x = w+4 = 8
we know that w=4 so we replace w by 4

Also since we know that n(B&#5198;C) = x+y = 4+y = 7
we know that y=3 so we replace y by 3

{{{drawing(300,300,-4,4,-5,4,

circle(0,-.5,2),
locate(-2,2,t),
locate(0,-2.7,C),
locate(-.3,-1,z),
locate(1.1,.4,3), 
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(-1.3,.5,4),
locate(0,2.5,2),
locate(2,2,v),
locate(-.2,1.1,4) )}}}

Now we are given that 

n(B-A) = v+y = 5

and since we know that v+y = 5, and y=3,
v+3 = 5 and so v=2.  So we replace v by 2 

{{{drawing(300,300,-4,4,-5,4,

circle(0,-.5,2),
locate(-2,2,t),
locate(0,-2.7,C),
locate(-.3,-1,z),
locate(1.1,.4,3), 
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(-1.3,.5,4),
locate(0,2.5,2),
locate(2,2,2),
locate(-.2,1.1,4) )}}}

Now we know how many elements there are in B,
n(B) = u+v+x+y = 2+2+4+3 = 11

So the cardinality of B is 11.

Now since n(B&#5196;C) = u+v+w+x+y+z = 20,
2+2+4+4+3+z = 20,
       15+z = 20
          z = 5

So now we replace z by 5

{{{drawing(300,300,-4,4,-5,4,

circle(0,-.5,2),
locate(-2,2,t),
locate(0,-2.7,C),
locate(-.3,-1,5),
locate(1.1,.4,3), 
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(-1.3,.5,4),
locate(0,2.5,2),
locate(2,2,2),
locate(-.2,1.1,4) )}}}

and now we know the cardinality of C, because

n(C) = w+x+y+z = 4+4+3+5 = 16

Since n(A&#5196;B&#5196;C) = t+u+v+w+x+y+z = 25,
t+2+2+4+4+3+5 = 25
         t+20 = 25
            t = 5
 
So we replace t by 5
{{{drawing(300,300,-4,4,-5,4,

circle(0,-.5,2),
locate(-2,2,5),
locate(0,-2.7,C),
locate(-.3,-1,5),
locate(1.1,.4,3), 
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(-1.3,.5,4),
locate(0,2.5,2),
locate(2,2,2),
locate(-.2,1.1,4) )}}}

So now we know that the cardinality of A is

n(A) = t+u+w+x = 5+2+4+4 = 15
 
Edwin</pre>