Question

Let A, B, and C be sets in a universal set U. We are given n(U) = 75, n(A) = 39, n(B) = 43, n(C) = 43, n(A ∩ B) = 27, n(A ∩ C) = 21, n(B ∩ C) = 28, n(A ∩ B ∩ CC) = 12. Find the following values.
(a) n(AC ∩ B ∩ C)

b) n(A ∩ BC ∩ CC)?

Answer 1

a.

$B\cap C=(B\cap C)\cap(A\cup A^c)=(A\cap B\cap C)\cup(A^c\cap B\cap C)$

which means

$n(B\cap C)=n(A\cap B\cap C)+n(A^c\cap B\cap C)$

$\implies n(A^c\cap B\cap C)=28-n(A\cap B\cap C)$

Similarly,

$n(A\cap B)=n(A\cap B\cap C)+n(A\cap B\cap C^c)$

$\implies n(A\cap B\cap C)=27-12=15$

Then

$n(A^c\cap B\cap C)=28-15=\boxed{13}$

b.

$n(A)=n(A\cap(B\cup C))+n(A\cap(B\cup C)^c)$

$n(A)=n(A\cap(B\cup C))+n(A\cap B^c\cap C^c)$

$\implies n(A\cap B^c\cap C^c)=39-n(A\cap(B\cup C))$

Note that $A\cap(B\cup C)=(A\cap B)\cup(A\cap C)$.

By the inclusion-exclusion principle,

$n(A\cap(B\cup C))=n(A\cap B)+n(A\cap C)-n((A\cap B)\cap(A\cap C))$

and $(A\cap B)\cap(A\cap C)=A\cap B\cap C$, so

$n(A\cap(B\cup C))=27+21-15=33$

Then

$n(A\cap B^c\cap C^c)=39-33=\boxed6$