The sum

Cardinality
We denote the size, or cardinality, of a finite set $S$ by: $\boxed{|S|}$

Note: $∣\emptyset∣ = 0$

For example:

Let $S = {a, b, c}$ hence $∣ S ∣ = 3$ . It is said that the cardinality of $S$ is 3, or $S$ has 3 elements.

If $A = {n \in N ∣ n is an odd number between 4 and 24}$ then $∣ A ∣ = 10$ .

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The sum rule
The sum rule states that if $A$ and $B$ are disjoint sets then $∣ A \cup B ∣ =$ $|A| + |B|$.

According to the sum rule, if $A_{1}, A_{2}, ..., A_{n}$ are $n$ parwise disjoint sets then $∣ A_{1} \cup A_{2} \cup ... \cup A_{n} ∣ =$ $|A_1| + |A_2| + ... + |A_n|$.
title: Example
A student can choose one project from project lists of three lecturers.
- John's list has 11.
- Bill's has 10.
- Eve's has 8.
 
No project occurs twice. How many possible projects are there to choose from?
- $A = \{ p | p \text{ is a project on John's list} \}$
- $B = \{ p | p \text{ is a project on Bill's list} \}$
- $C = \{ p | p \text{ is a project on Eve's list} \}$
 
Hence $|A \cup B \cup C| = |A| + |B| + |C| = 29$.
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Inclusion-exclusion principle

Inclusion-exclusion principle
The inclusion-exclusion principle states that $∣ A \cup B ∣ =$ $|A| + |B| - |A \cap B|$.

The inclusion-exclusion principle can be applied for three sets, by applying the same rule twice, $∣ A \cup (B \cup C) ∣$ , giving us: ?
$∣ A \cup B \cup C ∣ = ∣ A \cup (B \cup C) ∣ = ∣ A ∣ + ∣ B \cup C ∣ - ∣ A \cap (B \cup C) ∣ = ∣ A ∣ + ∣ B ∣ + ∣ C ∣ - ∣ B \cap C ∣ - ∣ A \cap (B \cup C) ∣ = ∣ A ∣ + ∣ B ∣ + ∣ C ∣ - ∣ B \cap C ∣ - ∣ (A \cap B) \cup (A \cap C) ∣ = ∣ A ∣ + ∣ B ∣ + ∣ C ∣ - ∣ B \cap C ∣ - (∣ A \cap B ∣ + ∣ A \cap C ∣ - ∣ (A \cap B) \cap (A \cap C) ∣) = ∣ A ∣ + ∣ B ∣ + ∣ C ∣ - ∣ B \cap C ∣ - ∣ A \cap C ∣ - ∣ A \cap C ∣ + ∣ A \cap B \cap C ∣$ Link to original