Answer:
[tex]n(A\ u\ B) = 10[/tex]
[tex]n(A'\ u\ C) = 10[/tex]
[tex]n(A\ n\ B)' = 6[/tex]
Step-by-step explanation:
Given
[tex]n(U) = 12[/tex]
[tex]n(A\ n\ B) =n(A\ n\ C) = n(B\ n\ C) =6[/tex]
[tex]n(A\ n\ B\ n\ C)=4[/tex]
[tex]n(A\ u\ B\ u\ C)=10[/tex]
Required
Solve a, b and c
There are several ways to solve this; the best is by using Venn diagram (see attachment for diagram)
Solving (a):
[tex]n(A\ u\ B)[/tex]
This is calculated as:
[tex]n(A\ u\ B) = n(A) + n(B) - n(A\ n\ B)[/tex]
From the attachment
[tex]n(A) = 0+2+4+2 = 8[/tex]
[tex]n(B) = 0+2+4+2 = 8[/tex]
[tex]n(A\ n\ B) = 4 +2 = 6[/tex]
So:
[tex]n(A\ u\ B) = 8 + 8 - 6[/tex]
[tex]n(A\ u\ B) = 10[/tex]
Solving (b):
[tex]n(A'\ u\ C)[/tex]
This is calculated as:
[tex]n(A'\ u\ C) = n(A') + n(C) -n(A'\ n\ C)[/tex]
From the attachment
[tex]n(A) = n(U) - n(A) = 12 - 8 = 4[/tex]
[tex]n(C) = 0+2+4+2 = 8[/tex]
[tex]n(A'\ n\ C) = 2[/tex]
So:
[tex]n(A'\ u\ C) = 4 + 8 - 2[/tex]
[tex]n(A'\ u\ C) = 10[/tex]
Solving (c):
[tex]n(A\ n\ B)'[/tex]
This is calculated as:
[tex]n(A\ n\ B)' = n(U) - n(A\ n\ B)[/tex]
[tex]n(A\ n\ B)' = 12- 6[/tex]
[tex]n(A\ n\ B)' = 6[/tex]
