Number of elements/cardinality of sets

Cardinality of sets

The cardinality of sets refers to the measure of a set's size, specifically the number of elements contained within the set. If A is a finite set with n elements, then the cardinality of A is n, denoted by $n(A)$. By definition, the cardinality of an empty set is zero since it contains no elements. When combining two or more sets, the cardinality of the union can be determined using a general formula that accounts for overlapping elements.

Cardinality of union of two sets

Consider two finite sets A and B within a universal set U, as shown in Figure 2.10.

Figure 2.10: Two enclosed sets in the universal set

The cardinality of the union of sets A and B is given by the formula:

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

This formula ensures that elements common to both sets are not counted twice.

Example 2.29

Given: $n(A) = 28$, $n(B) = 10$, and $n(A \cup B) = 33$. Find the cardinality of the intersection $n(A \cap B)$.

Solution: Using the union formula:

$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$

Rearranged to solve for $n(A \cap B)$:

$$n(A \cap B) = n(A) + n(B) - n(A \cup B) = 28 + 10 - 33 = 5$$

Therefore, the number of elements in both sets is $\boxed{5}$.

Representation and addition of cardinalities

If we denote the components of sets as:

• $n(A) = k + j$
• $n(B) = j + p$
• $n(A \cap B) = j$
• $n(A \cup B) = k + j + p$

Then the sum of cardinalities of A and B is:

$$n(A) + n(B) = (k + j) + (j + p) = k + 2j + p$$

Using the union formula:

$$n(A \cup B) + n(A \cap B) = (k + j + p) + j = k + 2j + p$$

This confirms the consistency of the general formula.

Application of cardinality in real contexts

In a bouquet of 44 flowers, there are 21 red flowers and 17 variegated flowers. If 7 flowers are neither red nor variegated, find how many flowers are both red and variegated.

Let:

• $R$ = set of red flowers
• $V$ = set of variegated flowers
• $x$ = number of flowers both red and variegated

Using the formula for the union of two sets:

$$n(R \cup V) = n(R) + n(V) - n(R \cap V)$$

We know that:

• Total flowers: $n(U) = 44$
• Neither red nor variegated: 7
• Thus, $n(R \cup V) = 44 - 7 = 37$

Substitute the values:

$$37 = 21 + 17 - x \implies x = 21 + 17 - 37 = 1$$

Therefore, there is 1 flower that is both red and variegated.

Cardinality of union of three sets

The formula for the union of two sets extends to three sets $A$, $B$, and $C$ as follows:

$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$$

This formula accounts for double counting and subtracts the overlaps accordingly, then adds back the intersection of all three sets.

Proof sketch

• By applying associative and distributive laws of sets:
• Express $n(A \cup B \cup C) = n((A \cup B) \cup C)$
• Use the union formula for two sets repeatedly and simplify.

Example

Given:

• $n(A) = 10$, $n(B) = 7$, $n(C) = 9$
• $n(A \cap B) = 4$, $n(B \cap C) = 3$, $n(A \cap C) = 3$
• $n(A \cap B \cap C) = 1$

Find $n(A \cup B \cup C)$.

Solution:

$$\begin{aligned} n(A \cup B \cup C) &= n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C) \\ &= 10 + 7 + 9 - 4 - 3 - 3 + 1 = 17 \end{aligned}$$

Hence, $n(A \cup B \cup C) = \boxed{17}$.

Example 2.32: Survey on machine operators and cleaners

A factory has 45 men and 45 women workers. Among them:

• 60 are machine operators
• 42 are cleaners
• 30 men are operators
• 30 men are cleaners
• 24 men are both operators and cleaners
• 12 women are neither operators nor cleaners

Find the number of workers who are both machine operators and cleaners.

Solution: Let $M$ = machine operators and $C$ = cleaners. Define $x$ as the number of women who are both operators and cleaners. From the data and Venn diagram logic:

$$(30 - x) + x + x + (12 - x) + 9 = 45$$

After simplification:

$$x = 9$$

The total number of workers both operators and cleaners is:

$$n(M \cap C) = 24 + 9 = 33$$

Therefore, 33 workers are both machine operators and cleaners.

Language speakers percentage problem

In a city:

• 85% speak English (E)
• 40% speak French (F)
• 20% speak Spanish (S)
• 32% speak English and French
• 13% speak English and Spanish
• 10% speak French and Spanish

Find the percentage who speak all three languages.

Solution: Let $x$ be the percentage speaking all three. Total percentage is 100, so using the formula:

$$\begin{aligned} 100 &= 85 + 40 + 20 - 32 - 13 - 10 + x \\ 100 &= 145 - 55 + x \\ 100 &= 90 + x \\ x &= 10 \end{aligned}$$

Hence, 10% speak all three languages.

Movie customers survey

A survey of 300 movie customers revealed interest in three movie types: Comedy (C), Story (S), and Action (A).

• $n(C) = 156$, $n(S) = 106$, $n(A) = 133$
• $n(C \cap S) = 53$, $n(S \cap A) = 41$, $n(C \cap A) = 87$
• $n(C \cap S \cap A) = 8$

Find:

1. Number who did not watch any of these movie types.
2. Number who watched comedy or action movies.
3. Number who watched story movies only.

Solution: (a) Customers who did not watch comedy, story, or action:

$$n(U) - n(C \cup S \cup A) = 300 - \bigl(156 + 106 + 133 - 53 - 41 - 87 + 8 \bigr) = 300 - 222 = 78$$

(b) Customers who watched comedy or action:

$$n(C \cup A) = n(C) + n(A) - n(C \cap A) = 156 + 133 - 87 = 202$$

(c) Customers who watched story only:

$$n(S) - n(S \cap A) - n(S \cap C) + n(C \cap S \cap A) = 106 - 41 - 53 + 8 = 20$$

Class of girls with various attributes

In a class:

• 15 girls like chocolate
• 5 girls are white
• 6 girls are intelligent
• Every white girl likes chocolate
• 3 intelligent girls do not like chocolate
• 2 girls are both white and intelligent

Find:

1. Draw a Venn diagram representing the data.
2. Find the total number of girls in the class.
3. Find the number of girls who are white but not intelligent.

Solution:

• Total number of girls $= 9 + 3 + 2 + 1 + 3 = 18$
• Number of white but not intelligent girls $= 3$