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)