คำตอบ

2014-02-08T20:11:44+07:00
สูตรเซตจำนวนสับเซตของA = 2^n(A)
n(P(A)) = 2^n(A) , n(P(P(A))) = 2^2^n(A)
P(A) U P(B) ⊂ P(A U B)
P(A) ∩ P(B) = P(A ∩ B)
A∩(B U C) = (A∩B) U (A∩C)
A U (B ∩ C) = (A U B) ∩ (A U C)
(A U B)' = A' ∩ B'
(A ∩ B)' = A' U B'
A - B = A ∩ B'
A - (B ∩ C) = (A - B) U (A - C)
A- (B U C) = (A - B) ∩ (A - C)
(A U B) - C = (A - C) U (B - C)
(A ∩ B) - C = (A - C) ∩ (B - C)
n(A U B) = n(A) + n(B) - n(A ∩ B)
n(A U B U C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(A ∩ C) - n(B ∩ C) + n(A ∩ B ∩ C)
ป.ล.เครื่องหมาย "^"คือยกกำลัง