SET THEORY-APPLICATION

Question 1: Set Operations and Venn Diagrams

In a class of 40 students:

 30 students take Mathematics (set MM).

 25 students take Physics (set PP).
 15 students take both Mathematics and Physics.

a) How many students take only Mathematics?

b) How many students take only Physics?
c) How many students take both Mathematics and Physics?
d) How many students take neither Mathematics nor Physics?

Solution: We can solve this using the principle of inclusion-exclusion and set operations.

 Let ∣M∣=30, ∣P∣=25, and ∣M∩P∣=15 (number of students taking both Mathematics and
Physics).

a) Students taking only Mathematics = ∣M∣−∣M∩P∣=30−15=15

b) Students taking only Physics = ∣P∣−∣M∩P∣=25−15=10
c) Students taking both Mathematics and Physics = ∣M∩P∣=15
d) Students taking neither Mathematics nor Physics = Total students - (Students taking

∣M∪P∣=∣M∣+∣P∣−∣M∩P∣=30+25−15=40
Mathematics or Physics)

Students taking neither = 40−∣M∪P∣=40−40=0

Question 2: Union of Sets

In a survey of 100 people:

 60 people like coffee (set CC).

 50 people like tea (set TT).
 30 people like both coffee and tea.

a) How many people like either coffee or tea?

b) How many people like only coffee?
c) How many people like neither coffee nor tea?

Solution: Use the inclusion-exclusion principle to solve the questions.

 Let ∣C∣=60, ∣T∣=50, and ∣C∩T∣=30

∣C∪T∣=∣C∣+∣T∣−∣C∩T∣=60+50−30=80
a) People who like either coffee or tea:

So, 80 people like either coffee or tea.

∣C∣−∣C∩T∣=60−30=30
b) People who like only coffee:

So, 30 people like only coffee.

c) People who like neither coffee nor tea:

Total people - people who like either coffee or tea = 100−80=20.
So, 20 people like neither.

Question 3: Subsets and Power set

Let A={1,2,3}A = \{1, 2, 3\}.

a) List all subsets of A.
b) How many elements are in the power set of A?

Solution:

 The set A= {1,2,3}

a) The subsets of A are:

 ∅ (empty set)
 {1}
 {2}
 {3}
 {1,2}
 {1,3}
 {2,3}
 {1,2,3}

∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}
So, the subsets of AA are:

b) The power set of a set with n elements has 2^n elements.

Since A has 3 elements, the power set has 2^3 = 8 elements.

Question 4: Complement of a Set

In a set of 200 students:

  120 students like sports (set SS).
 80 students like music (set MM).
 50 students like both sports and music.

a) How many students like only sports?

b) How many students like only music?
c) How many students like neither sports nor music?

Solution:

 Let ∣S∣=120|S| = 120, ∣M∣=80|M| = 80, and ∣S∩M∣=50

∣S∣−∣S∩M∣=120−50=70
a) Students who like only sports:

∣M∣−∣S∩M∣=80−50=30
b) Students who like only music:

c) Students who like neither sports nor music:

∣S∪M∣=∣S∣+∣M∣−∣S∩M∣=120+80−50=150
Total students - (Students who like sports or music)

Students who like neither = 200−150=50

Question 5: Cartesian Product

Let A={1,2} and B={a,b}

a) Find the Cartesian product A×B
b) How many elements are in A×B ?

Solution:

 The Cartesian product A×B is the set of all ordered pairs where the first element is from
A and the second is from B.

a) A×B= {(1, a), (1, b), (2, a), (,b)}

b) The number of elements in A×B is ∣A∣×∣B∣=2×2=4

Three Sets and Venn Diagrams

More examples

Question 1: Three Sets and Venn Diagrams

In a survey of 150 people:

 100 people like Apple (set AA).

 80 people like Banana (set BB).
 60 people like Cherry (set CC).
 50 people like both Apple and Banana.
 40 people like both Apple and Cherry.
 30 people like both Banana and Cherry.
 20 people like all three fruits: Apple, Banana, and Cherry.

a) How many people like Apple but not Banana or Cherry?

b) How many people like at least one of the three fruits?
c) How many people like only one of the three fruits?

Solution:

Let’s use the principle of inclusion-exclusion to help us.

∣A∣=100, ∣B∣| = 80, ∣C∣=60|

∣A∩B∣ = 50, ∣A∩C∣=40, ∣B∩C∣=30


∣A∩B∩C∣=20



and A∩C : ∣A∩B∪A∩C∣=∣A∩B∣+∣A∩C∣−∣A∩B∩C∣=50+40−20=70. So:

a) To find how many people like Using the inclusion-exclusion principle on the union of A∩B

∣A∖(B∪C)∣=100−70=30. Therefore, 30 people like Apple but not Banana or Cherry.

∣A∪B∪C∣=∣A∣+∣B∣+∣C∣−∣A∩B∣−∣A∩C∣−∣B∩C∣+∣A∩B∩C∣|. Substituting the values:

b) To find how many people like at least one fruit, we use:

∣A∪B∪C∣=100+80+60−50−40−30+20=140. Therefore, 140 people like at least one of the three

fruits.

c) To find how many people like only one of the three fruits, we subtract the number of people
who like two or three fruits from the total in each set.

∣A∖(B∪C)∣=30|
 People who like only Apple:

∣B∖(A∪C)∣=∣B∣−(∣A∩B∣+∣B∩C∣−∣A∩B∩C∣)=80−(50+30−20)=20.
 People who like only Banana:

∣C∖(A∪B)∣=∣C∣−(∣A∩C∣+∣B∩C∣−∣A∩B∩C∣)=60−(40+30−20)=10
 People who like only Cherry:
Thus, the number of people who like only one fruit is: 30+20+10=6030 + 20 + 10 = 60

Question 2: Four Sets and Inclusion-Exclusion

In a community of 200 people, the following information is known:

 120 people like Football (set FF).

 90 people like Basketball (set BB).
 80 people like Baseball (set LL).
 70 people like Tennis (set TT).
 60 people like both Football and Basketball.
 50 people like both Football and Baseball.
 40 people like both Football and Tennis.
 30 people like both Basketball and Baseball.
 20 people like both Basketball and Tennis.
 10 people like both Baseball and Tennis.
 15 people like all four sports.

a) How many people like exactly two sports?

b) How many people like at least three sports?
c) How many people like none of the sports?

Solution:

Let’s again use the inclusion-exclusion principle.

Let’s define the given sets and the relationships between them:

∣F∣=120|F| = 120, ∣B∣=90|B| = 90, ∣L∣=80|L| = 80, ∣T∣=70|T| = 70

∣F∩B∣=60, ∣F∩L∣=50, ∣F∩T∣=40


∣B∩L∣=30, ∣B∩T∣=20, ∣L∩T∣=10



∣F∩B∩L∩T∣=15



a) To find how many people like exactly two sports, we calculate the number of people in each
pairwise intersection, excluding those who like three or four sports.

For example:

∣F∩B∣−∣F∩B∩L∩T∣=60−15=45
 People who like exactly Football and Basketball:

∣F∩L∣−∣F∩B∩L∩T∣=50−15=35
 People who like exactly Football and Baseball:

∣F∩T∣−∣F∩B∩L∩T∣=40−15=25
 People who like exactly Football and Tennis:
 ∣B∩L∣−∣F∩B∩L∩T∣=30−15=15
 People who like exactly Basketball and Baseball:

∣B∩T∣−∣F∩B∩L∩T∣=20−15=5
 People who like exactly Basketball and Tennis:

∣L∩T∣−∣F∩B∩L∩T∣=10−15=0(This set has no people).

 People who like exactly Baseball and Tennis:

So the total number of people who like exactly two sports is: 45+35+25+15+5+0=12545 + 35 +
25 + 15 + 5 + 0 = 125

b) To find how many people like at least three sports, we calculate the number of people who
like exactly three sports or all four sports.

People who like exactly three sports are in the triple intersections minus those who like all four
sports:

∣F∩B∩L∣−∣F∩B∩L∩T∣=15−15=0
 People who like exactly Football, Basketball, and Baseball:

∣F∩B∩T∣−∣F∩B∩L∩T∣=10−15=0
 People who like exactly Football, Basketball, and Tennis:

∣F∩L∩T∣−∣F∩B∩L∩T∣=10−15=0
 People who like exactly Football, Baseball, and Tennis:

∣B∩L∩T∣−∣F∩B∩L∩T∣=10−15=0
 People who like exactly Basketball, Baseball, and Tennis:

The number of people who like at least three sports is 15 (those who like all four sports).

c) To find how many people like none of the sports, we use:

People who like none=Total people−∣F∪B∪L∪T∣

Using inclusion-exclusion to find ∣F∪B∪L∪T∣|:

∣F∪B∪L∪T∣=∣F∣+∣B∣+∣L∣+∣T∣−∣F∩B∣−∣F∩L∣−∣F∩T∣−∣B∩L∣−∣B∩T∣−∣L∩T∣+∣F∩B∩L∣+∣F∩B∩T∣+∣
F∩L∩T∣+∣B∩L∩T∣−∣F∩B∩L∩T∣|

Substituting the values:

∣F∪B∪L∪T∣=120+90+80+70−60−50−40−30−20−10+15+10+10+10−15=160

So, the number of people who like none is: 200−160=40200 - 160 = 40