SOLUTION: I am working on cardinality of sets and the problem is: n(AᑎB)=6 n(BᑎC)=7 n(AᑎC)=8 n(AᑌBᑌC)=25 n(AᑎBᑎC)=4 n(B-A)=5 n(BᑌC)=2

 
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.



n(A) = t+u+w+x
n(B) = u+v+x+y
n(C) = w+x+y+z
n(AᑎB) = u+x = 6
n(AᑎC) = w+x = 8
n(BᑎC) = x+y = 7
n(AᑎBᑎC) = x = 4
n(AᑌB) = t+u+v+w+x+y
n(AᑌC) = t+u+w+x+y+z
n(BᑌC) = u+v+w+x+y+z = 20
n(AᑌBᑌC) = 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(AᑎBᑎC)= x = 4

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



n(AᑎB) = u+x = 6
n(AᑎC) = w+x = 8
n(BᑎC) = x+y = 7

Since we know that n(AᑎB) = u+x = u+4 = 6
we know that u=2 so we replace u by 2

Also since we know that n(AᑎC) = w+x = w+4 = 8
we know that w=4 so we replace w by 4

Also since we know that n(BᑎC) = x+y = 4+y = 7
we know that y=3 so we replace y by 3



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 



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(BᑌC) = 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



and now we know the cardinality of C, because

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

Since n(AᑌBᑌC) = 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


So now we know that the cardinality of A is

n(A) = t+u+w+x = 5+2+4+4 = 15
 
Edwin