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Some Algebraic Implications of Inequalities

if A,B,C are ...

Question

if A,B,C are three finite sets such that n(A)=27,n(B)=23,n(C)=25 AND n(A intersection B)=14 ,n(BINTERSECTION C)=9,n(C INTERSECTION A)=10 AND n(A inter B inter C)=6 ALSO n(U)=80
FIND;n(A INTER B' INTER C')
n(B INTER A'C')
n(C INTER A' INTER B')

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Solution

$n (A) = 27, n (B) = 23, n (C) = 25, n (A \cap B) = 14, n (B \cap C) = 9, n (C \cap A) = 10, n (A \cap B \cap C) = 6 and n (U) = 80 (1) n (A \cap B' \cap C') = n (A \cap (B \cup C)') = n (A) - n (A \cap (B \cup C)) (Since n (A \cap B') = n (A) - n (A \cap B)) = n (A) - (n (A \cap B) \cup n (A \cap C)) = n (A) - (n (A \cap B) + n (A \cap C) - n (A \cap B \cap C)) = 27 - (14 + 10 - 6) = 27 - 18 = 9 ANS . . . (2) n (B \cap A' \cap C') = n (B \cap (A \cup C)') = n (B) - n (B \cap (A \cup C)) (Since n (A \cap B') = n (A) - n (A \cap B)) = n (B) - (n (A \cap B) \cup n (B \cap C)) = n (B) - (n (A \cap B) + n (B \cap C) - n (A \cap B \cap C)) = 23 - (14 + 9 - 6) = 23 - 17 = 6 ANS . . . (3) n (C \cap A' \cap B') = = n (C \cap (A \cup B)') = n (C) - n (C \cap (A \cup B)) (Since n (A \cap B') = n (A) - n (A \cap B)) = n (C) - (n (C \cap A) \cup n (B \cap C)) = n (C) - (n (A \cap C) + n (B \cap C) - n (A \cap B \cap C)) = 25 - (10 + 9 - 6) = 27 - 13 = 14 ANS . . .$

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