Set Theory (Part II) Counting Principles for the Union and Intersection of Sets.

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Set Theory (Part II)

Counting Principles for the Union and Intersection of

Sets

In some cases, the number of elements that exist in a set is

needed.With simple sets, direct

counting is the quickest way.

For example: Given any class, a student can either pass or fail.

(These sets are called “mutually exclusive”)

If 3 students fail, and 22 students pass, how many students are there in the class?

3 + 22 = 25

Not all calculations involve ME sets

For example: Consider a group of teachers and classes.

12 math teachers8 physics teachers

3 teach bothHow many teachers are there?

12: math8: physics

3: both

Can we just add them up? 12 + 8 + 3 = 23?

NO WAY!!!

Try drawing a Venn Diagram

U

U = all the teachers in the school

Begin with the overlap: 3 people like both

M = math (12)

P = physics (8)

3

M P

9 5

U

Add up all the individual spaces:

9 + 3 + 5 = 17

3

M P

9 5

Can we get 17 from the original numbers?

12 8 3 17+ - =

In general:

Algebraically: n(A U B) =

n(A) + n(B) – n(A B)U

Consider a situation with 3 distinguishing features.

In a school, the following is true:

30 students are on the football team

15 are on the hockey team

25 are on the track team

8 are on football and hockey

6 are on hockey and track

12 are on football and track

4 are on all 3 teams

How many students are there in total?

U = students at the schoolF = football playersH = hockey playersT = track members

For the Venn Diagram, begin with the center and work your way out…

U

T

F H

In a school, the following is true:

30 students are on the football team

15 are on the hockey team

25 are on the track team

8 are on football and hockey

6 are on hockey and track

12 are on football and track

4 are on all 3 teams

How many students are involved on all 3 teams?

U

T

F H

4

In a school, the following is true:

30 students are on the football team

15 are on the hockey team

25 are on the track team

8 are on football and hockey

6 are on hockey and track

12 are on football and track

4 are on all 3 teams

How many students are involved on all 3 teams?

U

T

F H

48

In a school, the following is true:

30 students are on the football team

15 are on the hockey team

25 are on the track team

8 are on football and hockey

6 are on hockey and track

12 are on football and track

4 are on all 3 teams

How many students are involved on all 3 teams?

U

T

F H

48 2

In a school, the following is true:

30 students are on the football team

15 are on the hockey team

25 are on the track team

8 are on football and hockey

6 are on hockey and track

12 are on football and track

4 are on all 3 teams

How many students are involved on all 3 teams?

U

T

F H

48 2

4

In a school, the following is true:

30 students are on the football team

15 are on the hockey team

25 are on the track team

8 are on football and hockey

6 are on hockey and track

12 are on football and track

4 are on all 3 teams

How many students are involved on all 3 teams?

U

T

F H

48 2

4

11

In a school, the following is true:

30 students are on the football team

15 are on the hockey team

25 are on the track team

8 are on football and hockey

6 are on hockey and track

12 are on football and track

4 are on all 3 teams

How many students are involved on all 3 teams?

U

T

F H

48 2

4

11

5

In a school, the following is true:

30 students are on the football team

15 are on the hockey team

25 are on the track team

8 are on football and hockey

6 are on hockey and track

12 are on football and track

4 are on all 3 teams

How many students are involved on all 3 teams?

U

T

F H

48 2

4

11

514

Add up all the numbers = 48

Worksheet

Try it with our numbersThe number of students involved is:

30 + 15 + 25 – 8 – 6 – 12 + 4 = 48

In general:

n(A U B U C) = n(A) + n(B) + n(C)

- n(A B) – n(A C) – n(B C)

+ n(A B C)

U U U

U U

U

Start by adding each subset and track the overlap … (on board)