The cardinality of set A is defined as the number of elements in the set A and is denoted by n(A).
For example, if A = {a,b,c,d,e} then cardinality of set A i.e.n(A) = 5

Let A and B are two subsets of a universal set U. Their relation can be shown in Venn-diagram as:

$$ n(A) = n_o( A) + n(A \cap B)$$

$$\text{or,}\: n(A) - n (A \cap B)= n_o(A)$$

$$ n(B) = n_o(B) + n(A \cap B)$$

$$\text {or,}\: n(B) - n(A \cap B) = n_o(B)$$

Also,

\begin{align*} n(A∪B) &= n_o(A) + n(A∩B) + n_o(B)\\ n(A∪B) &= n(A) - n(A∩B) + n(A∩B) + n(B) - n(A∩B)\\ n(A∪B) &= n(A) + n(B)- n(A∩B)\\ \therefore n(A∪B) &= n(A) + n(B) - n(A∩B)\\ \end{align*}

If A and B are disjoint sets then:

\(n(A \cap B) = 0, n(A \cup B) =n(A) + n(B)\)

Again,

\(n(U) = n(A \cup B) + n(\overline {A\cup B)}\)

If \(n(\overline {A \cup B)}\)=0, then \( n(U) = n(A \cup B)\)

Problems involving three sets

Let A, B and C are three non-empty and intersecting sets, then:
\(n(A \cup B \cup C) = n(A) + n(B) +n(C) - n(A \cap B) -n(B \cap C) -n(C \cap A) +n(A \cap B \cap C).\)

In Venn-diagram

\(n(A)\) = Number of elements in set A.

\(n(B)\) = Number of elements in set B.

\(n(C)\)=Number of element in set C.

\(n_o(A)\) = Number of elements in set A only.

\(n_o(B)\) = Number of elements in set B only.

\(n_o(C)\) = Number of elements in set C only.

\(n_o(A \cap B)\) = Number of elements in set A and B only.

\(n_o(B \cap C)\) = Number of elements in set B and C only.

\(n_o(C \cap A)\) = Number of elements in set A and C only.

\(n(A \cap B \cap C)\) = Number of elements in set A, B and C.

From the Venn-diagram

\begin{align*} n(A \cup B \cup C) &= n_o(A) +n_o(B) +n_o(C) +n_o(A \cap B) +n_o(B \cap C) +n_o(C \cap A) + n(A \cap B \cap C)\\ &= n(A) - n_o(A \cap B) - n_o(C \cap A) - n(A \cap B \cap C) + n(B) - n_o(B \cap C) - n_o(C\cap B) - n(A \cap B \cap C)
+ n(C) - n_o(A \cap C) - n_o(B \cap C) - n(A \cap B \cap C) + n_o(A \cap B) +n_o(B \cap C) +n_o(C \cap A) + n(A \cap B \cap C)\\ &= n(A) + n(B) + n(C) - [n_o(A \cap B) +n(A \cap B \cap C)] - [n_o(A \cap B) +n(A \cap B \cap C)] - [n_o(B \cap C) +n(A \cap B \cap C)] - [n_o(C \cap A) +n(A \cap B \cap C)]+n(A \cap B \cap C)\\ &= n(A) + n(B) + n(C) - n(A \cap B) -n(B \cap C) -n(A \cap C) +n(A \cap B \cap C)\\ \end{align*}

$$\boxed{\therefore (A \cup B \cup C)= n(A) + n(B) + n(C) - n(A \cap B) -n(B \cap C) -n(C \cap A) +n(A \cap B \cap C)} $$

If A, B and C are disjoint sets,

\(n(A \cup B \cup C) = n(A) + n(B) + n(C)\)