If \( A, B \) and \( C \) are any three sets, then
A. \( A-(B \cap C)=(A \cap B)-(A \cap C) \) B. \( A-(B \cup C)=(A-B) \cap(A-C) \) C. \( n(A-B)=n(A)-n(A \cap B) \) D. \( A \cap(B-C)=(A \cap B) \cap(A-C) \) Choose the most appropriate answer from the options given below:Statement A:
\[
A - (B \cap C) = (A \cap B) - (A \cap C)
\]
Expanding the left-hand side:
\[
A - (B \cap C) = A \cap (B \cap C)^c
\]
Using De Morgan’s law:
\[
(B \cap C)^c = B^c \cup C^c
\]
So,
\[
A \cap (B^c \cup C^c) = (A \cap B^c) \cup (A \cap C^c)
\]
which does not match the right-hand side.
Thus, Statement A is incorrect.
Statement B:
\[
A - (B \cup C) = (A - B) \cap (A - C)
\]
Expanding both sides:
\[
A - (B \cup C) = A \cap (B \cup C)^c
\]
Using De Morgan’s law:
\[
(B \cup C)^c = B^c \cap C^c
\]
So,
\[
A \cap (B^c \cap C^c) = (A \cap B^c) \cap (A \cap C^c)
\]
Since:
\[
A - B = A \cap B^c, \quad A - C = A \cap C^c
\]
We get:
\[
(A - B) \cap (A - C) = (A \cap B^c) \cap (A \cap C^c)
\]
which matches the left-hand side, so Statement B is correct.
Statement C:
\[
n(A - B) = n(A) - n(A \cap B)
\]
From set theory, the number of elements in \( A - B \) is:
\[
n(A - B) = n(A \cap B^c)
\]
Since:
\[
n(A) = n(A \cap B) + n(A \cap B^c)
\]
Rearranging:
\[
n(A \cap B^c) = n(A) - n(A \cap B)
\]
which confirms Statement C is correct.
Statement D:
\[
A \cap (B - C) = (A \cap B) \cap (A - C)
\]
Expanding the left-hand side:
\[
B - C = B \cap C^c
\]
So:
\[
A \cap (B - C) = A \cap (B \cap C^c)
\]
Rearrange:
\[
(A \cap B) \cap C^c
\]
Since:
\[
A - C = A \cap C^c
\]
We get:
\[
(A \cap B) \cap (A - C)
\]
which confirms Statement D is correct.
Final Answer:
A is incorrect.
B, C, and D are correct.
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