SOLUTION: Given n(U) = 26, n(A) = 11. n(B) = 12. and n(A ∩ B) = 8, find; a. n(A U B) b. n(B, but not A) I don't understand the concept of solving these. Please explain in a little

a) 



If you draw a Venn diagram, you will have two overlapping circles, one for A and one for B.  The region where they overlap is A ∩ B.   There are 11 elements in  A, 12 in B, and 8 in A ∩ B (so, some of the elements in A are in A ∩ B, and likewise for B).   Thus n(A U B) = 12+11 -8 = 15 elements.   The subtraction of 8 (= n(A ∩ B) ) is to account for the fact that adding n(A) and n(B) counts the intersection twice.



n(A U B) = n(A) + n(B) - n(A ∩ B)





Concrete Example:

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

Say  A = { a, b, c, d, e, f, g }

and B = { a, e, i, o, u, y }



then  A ∩ B = { a, e }

and   A U B = { a, b, c, d, e, f, g, i, o, u, y }



n(A) = 7  

n(B) = 6



n(A ∩ B) = 2

n(A U B) = n(A) + n(B) - n(A ∩ B) = 7 + 6 - 2  = 11  (which is easily verified by counting the number of elements in A U B  directly)







b) 

 n(B, but not A) is the same as  n(B) - n(A ∩ B) = 12 - 8 = 4







Referring to the example: 

    n(B) - n(A ∩ B) = 6 - 2 = 4  (4 elements of B are ONLY in B, and those elements are {i,o,u,y} )