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: Find total elements in $U$

$n(U) = 12$ (count: 1,2,3,4,5,6,7,a,b,c,d,e)

Step2: Find $n(A)$ for part (a)

$n(A) = 4$ (count: 1,2,a,e)

Step3: Calculate $n(A^C)$ for (a)

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

Step4: Find $n(B)$ for part (b)

$n(B) = 7$ (count:1,2,3,4,a,b,c)

Step5: Calculate $n(B^C)$

$n(B^C) = n(U) - n(B) = 12 - 7 = 5$ (elements:5,6,7,d,e)

Step6: Find $A \cap B^C$ for (b)

$A \cap B^C = \{e\}$, so $n(A \cap B^C) = 1$

Step7: Calculate $n(A \cup B^C)$ for (c)

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

Step8: Calculate $n(A^C \cap B^C)$ for (d)

Use 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 $n(A \cap B)=3$
$n(A \cup B) = 4 + 7 - 3 = 8$
$n(A^C \cap B^C) = n(U) - n(A \cup B) = 12 - 8 = 4$

Answer:

(a) $\boldsymbol{8}$
(b) $\boldsymbol{1}$
(c) $\boldsymbol{8}$
(d) $\boldsymbol{4}$