Question

let ( u = {1, 2, 3, 4, 5, 6, 7, a, b, c, d, e} ). if ( a = {1, 2, a, e} ) and ( b = {1, 2, 3, 4, a, b, c} ), find the following.
(a) ( n(a^c) )
(b) ( n(a cap b^c) )
(c) ( n(a cup b^c) )
(d) ( n(a^c cap b^c) )

Explanation:

Step1: Define sets and counts

$U = \{1,2,3,4,5,6,7,a,b,c,d,e\}$, so $n(U)=12$
$A = \{1,2,a,e\}$, so $n(A)=4$
$B = \{1,2,3,4,a,b,c\}$, so $n(B)=7$

Step2: Solve (a): Find $n(A^C)$

Complement formula: $n(A^C)=n(U)-n(A)$
$n(A^C)=12-4=8$

Step3: Solve (b): Find $n(A \cap B^C)$

First find $n(B^C)=n(U)-n(B)=12-7=5$
$A \cap B^C$ is elements in $A$ not in $B$: $\{e\}$
$n(A \cap B^C)=1$

Step4: Solve (c): Find $n(A \cup B^C)$

Use union formula: $n(A \cup B^C)=n(A)+n(B^C)-n(A \cap B^C)$
$n(A \cup B^C)=4+5-1=8$

Step5: Solve (d): Find $n(A^C \cap B^C)$

De Morgan's Law: $A^C \cap B^C=(A \cup B)^C$
First find $n(A \cup B)=n(A)+n(B)-n(A \cap B)$
$A \cap B = \{1,2,a\}$, so $n(A \cap B)=3$
$n(A \cup B)=4+7-3=8$
$n((A \cup B)^C)=n(U)-n(A \cup B)=12-8=4$

Answer:

(a) 8
(b) 1
(c) 8
(d) 4