If A, B, and C are three sets such that A∩B = A∩C and A∪B = A∪C, then A=B, A=C, B=C, or A∩B=∅?

A∪B=A∪C ⇒n(A∪B)=n(A∪C) ⇒n(A)+n(B)−n(A∩B)=n(A)+n(C)−n(A∩C)
Since A∩B=A∩C, we have n(B)=n(C). However, this does not imply B=C. For example, consider A={1,2}, B={2,3}, C={2,4}. Then A∩B={2}, A∩C={2}, A∪B={1,2,3}, A∪C={1,2,4}. Thus n(B)=n(C)=2, but B≠C. However, if we assume that A∩B=A∩C=∅, then we have n(A)+n(B)=n(A)+n(C) which implies n(B)=n(C). Even with this assumption, this doesn't necessarily imply B=C. Let's consider the case when A∩B = A∩C. We use the principle of inclusion-exclusion. A∪B = A∪C implies that |A∪B| = |A∪C|. Using the inclusion-exclusion principle, we get:
|A∪B| = |A| + |B| - |A∩B|
|A∪C| = |A| + |C| - |A∩C|
Since |A∪B| = |A∪C| and |A∩B| = |A∩C|, we have:
|A| + |B| - |A∩B| = |A| + |C| - |A∩C|
|B| = |C|
This only shows that B and C have the same cardinality (number of elements), not that they are equal as sets. Therefore, none of the options A=B, A=C, or A∩B=∅ are necessarily true. The correct conclusion is B=C.