For 3 sets A , B & C
n(A) = Number of elements of set A
n(B) = Number of elements of set B
n(C) = Number of elements of set C
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(A ∩ C) + n(A ∩ B ∩ C)
Proof of n(A ∪ B ∪ C) Formula
We know that
P(E ∪ F) = P(E) + P(F) − P(E ∩ F).
Putting E = B, F = (B ∪ C)
P(A ∪ (B ∪ C)) = P(A) + P(B ∪ C) − P(A ∩ (B ∪ C)),
And also
P(A ∩ (B ∪ C)) = P((A ∩ B) ∪ (A ∩ C))
= P(A ∩ B) + P(A ∩ C) − P(A ∩ B ∩ A ∩ C)
= P(A ∩ B) + P(A ∩ C) − P(A ∩ B ∩ C)
and
P(B ∪ C) = P(B) + P(C) − P(B ∩ C).
Substituting the 2 nd and 3 rd equations into the first we get
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(B ∩ C) − P(A ∩ B) − P(A ∩ C) + P(A ∩ B ∩ C)
