For any three sets A, B and C the set $(A∪B∪C)∩(A∩B'∩C')'∩C'$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $B∩C'$
We have,
$(A∩B'∩C')' = A'∪B∪C$
$∴(A∪B∪C)∩(A∩B'∩C')'$
$=(A∪B∪C)∩(A'∪B∪C)$
$=(A∩A')∪(B∪C)$ [By distributivity ∪ of over ∩]
$=\phi∪(B∪C) = B∪C$
Hence,
$(A∪B∪C)∩(A∩B'C')' ∪C'$
$=(B∪C)∩C' = (B∩C')∪(C∩C') = B∩C'$