Math 1300: Section 7- 3 Basic Counting Principles

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Addition PrincipleVenn Diagrams

Multiplication Principle

Math 1300 Finite MathematicsSection 7-3: Basic Counting Principles

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

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Example: If enrollment in a section of Math 1300 consists of 13males and 15 females, then it is clear that there are a total of 28students in the class.

To represent this in terms of set operations, we would firstassign names to the sets. Let

M = set of male students in the sectionF = set of female students in the section

Notice that M ∪ F is the set of all students in the class, and thatM ∩ F = ∅. The total number of students in the class is thenrepresented by n(M ∪ F ), and we have

n(M ∪ F ) = n(M) + n(F )

= 13 + 15 = 28.


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n(M ∪ F ) = n(M) + n(F )

= 13 + 15 = 28.


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Notice that M ∪ F is the set of all students in the class, and thatM ∩ F = ∅.

The total number of students in the class is thenrepresented by n(M ∪ F ), and we have

n(M ∪ F ) = n(M) + n(F )

= 13 + 15 = 28.


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n(M ∪ F ) = n(M) + n(F )

= 13 + 15 = 28.


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Example: Suppose we are told that an economics classconsists of 22 business majors and 16 journalism majors.

Let B represent the set of business majors and J represent theset of journalism majors.

Then n(B) = 22 and n(J) = 16, and B ∪ J is the set of allstudents in the class.

Can we conclude that n(B ∪ J) = 22 + 16 = 38?

No! This would double count double majors.

The set B ∩ J represents the set of students majoring in bothbusiness and journalism. If B ∩ J 6= ∅, then we must avoidcounting these students twice.


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If we had, say, 7 double majors in the class, then

n(B ∩ J) = 7

And the correct count would be

n(B ∪ J) = n(B) + n(J)− n(B ∩ J)= 22 + 16− 7= 31


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If we had, say, 7 double majors in the class, then

n(B ∩ J) = 7

And the correct count would be

n(B ∪ J) = n(B) + n(J)− n(B ∩ J)= 22 + 16− 7= 31


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Theorem (Addition Principle (For Counting))For any two sets A and B,

n(A ∪ B) = n(A) + n(B)− n(A ∩ B)

If A and B are disjoint, then

n(A ∪ B) = n(A) + n(B)


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Example: A marketing survey of a group of kids indicated that25 owned a Nintendo DS and 15 owned a Wii. If 10 kids hadboth a DS and a Wii, how many kids interviewed have a DS ora Wii?

Let D represent the set of kids with a DS, and let W representthe set of kids with a Wii.

n(D ∪W ) = n(D) + n(W )− n(D ∩W )

n(D ∪W ) = 25 + 15− 10 = 30

Number of kids with a DS or a Wii: 30.


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n(D ∪W ) = n(D) + n(W )− n(D ∩W )

n(D ∪W ) = 25 + 15− 10 = 30



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n(D ∪W ) = n(D) + n(W )− n(D ∩W )

n(D ∪W ) = 25 + 15− 10 = 30



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n(D ∪W ) = n(D) + n(W )− n(D ∩W )

n(D ∪W ) = 25 + 15− 10 = 30



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n(D ∪W ) = n(D) + n(W )− n(D ∩W )

n(D ∪W ) = 25 + 15− 10 = 30



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In problems which involve more than two sets or which involvecomplements of sets, it is often helpful to draw a VennDiagram.

Example: In a certain class, there are 23 majors in Psychology,16 majors in English and 7 students who are majoring in bothPsychology and English. If there are 50 students in the class,how many students are majoring in neither of these subjects?How many students are majoring in Psychology alone?

Let P represent the set of Psychology majors and let Erepresent the set of English majors.


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Then n(P) = 23, n(E) = 16, and n(P ∩ E) = 7. We are alsogiven that there are 50 students in the class, so n(U) = 50.Now we draw a Venn Diagram:

n(U) = 50

P E

716 9

18


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Then n(P) = 23, n(E) = 16, and n(P ∩ E) = 7. We are alsogiven that there are 50 students in the class, so n(U) = 50.Now we draw a Venn Diagram:n(U) = 50

P E

716 9

18


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P E

7

16 9

18


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P E

716

9

18


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P E

716 9

18


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P E

716 9

18


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Example: A survey of 100 college faculty who exerciseregularly found that 45 jog, 30 swim, 20 cycle, 6 jog and swim,1 jogs and cycles, 5 swim and cycle, and 1 does all three. Howmany of the faculty members do not do any of these threeactivities? How many just jog?


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Let A and B be sets and A ⊂ U, B ⊂ U,

n(A′) = n(U)− n(A)DeMorgan’s Laws

(A ∪ B)′ = A′ ∩ B′

(A ∩ B)′ = A′ ∪ B′


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Let A and B be sets and A ⊂ U, B ⊂ U,n(A′) = n(U)− n(A)

DeMorgan’s Laws(A ∪ B)′ = A′ ∩ B′

(A ∩ B)′ = A′ ∪ B′


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Let A and B be sets and A ⊂ U, B ⊂ U,n(A′) = n(U)− n(A)DeMorgan’s Laws

(A ∪ B)′ = A′ ∩ B′

(A ∩ B)′ = A′ ∪ B′


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Let A and B be sets and A ⊂ U, B ⊂ U,n(A′) = n(U)− n(A)DeMorgan’s Laws

(A ∪ B)′ = A′ ∩ B′

(A ∩ B)′ = A′ ∪ B′


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Example: Suppose that A and B are sets with n(A′) = 81,n(B′) = 90, n(A′ ∩ B′) = 63, and n(U) = 180. Determine thenumber of elements in each of the four disjoint subsets in thefollowing Venn diagram.

UA B

63

7227 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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UA B

63

7227 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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UA B

63

7227 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108

n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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UA B

63

7227 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72

n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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UA B

63

7227 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99

n(B) = 180− n(B′) = 90


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UA B

63

7227 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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UA B

63

72

27 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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UA B

63

72

27 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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UA B

63

7227

18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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UA B

63

7227 18

n(A′ ∪ B′) = n(A′) + n(B′)− n(A′ ∩ B′)

= 81 + 90− 63 = 108n(A ∩ B) = 180− n(A′ ∪ B′)

= 180− 108 = 72n(A) = 180− n(A′) = 99n(B) = 180− n(B′) = 90


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Example: Suppose you have 4 pairs of pants in your closet, 3different shirts and 2 pairs of shoes. How many different wayscan you choose an outfit consisting of one pair of pants, oneshirt and one pair of shoes?

Let’s consider this as a sequence of operations:

O1 Choose a pair of pantsO2 Choose a shirtO3 Choose a pair of shoes

Now for each operation, there is a specified number of ways toperform this operation:Operation Number of Ways

O1 N1 = 4O2 N2 = 3O3 N3 = 2


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Let’s consider this as a sequence of operations:

O1 Choose a pair of pantsO2 Choose a shirtO3 Choose a pair of shoes


O1 N1 = 4O2 N2 = 3O3 N3 = 2


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Let’s consider this as a sequence of operations:O1 Choose a pair of pants

O2 Choose a shirtO3 Choose a pair of shoes


O1 N1 = 4O2 N2 = 3O3 N3 = 2


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Let’s consider this as a sequence of operations:O1 Choose a pair of pantsO2 Choose a shirt

O3 Choose a pair of shoes


O1 N1 = 4O2 N2 = 3O3 N3 = 2


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Let’s consider this as a sequence of operations:O1 Choose a pair of pantsO2 Choose a shirtO3 Choose a pair of shoes


O1 N1 = 4O2 N2 = 3O3 N3 = 2


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Let’s consider this as a sequence of operations:O1 Choose a pair of pantsO2 Choose a shirtO3 Choose a pair of shoes


O1 N1 = 4O2 N2 = 3O3 N3 = 2


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So we have

i Oi Ni1 O1 42 O2 33 O3 2

Then we can draw a tree diagram to see that there areN1 · N2 · N3 = 4(3)(2) = 24 different outfits.


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So we have

i Oi Ni1 O1 42 O2 33 O3 2

Then we can draw a tree diagram to see that there areN1 · N2 · N3 = 4(3)(2) = 24 different outfits.


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Theorem (Multiplication Principle)If two operations O1 and O2 are performed in order, with N1possible outcomes for the first operation and N2 possibleoutcomes for the second operation, then there are

N1 · N2

possible combined outcomes for the first operation followed bythe second.


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Theorem (Generalized Multiplication Principle)In general, if n operations O1,O2, · · · ,On are performed inorder, with possible number of outcomes N1,N2, . . . ,Nn,respectively, then there are

N1 · N2 · · ·Nn

possible combined outcomes of the operations performed in thegiven order.


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Example: A fair 6-sided die is rolled 5 times, and each time theresulting sequence of 5 numbers is recorded.

(a) How many different sequences are possible?

6× 6× 6× 6× 6 = 7776

(b) How many different sequences are possible if all numbersexcept the first must be odd?

6× 3× 3× 3× 3 = 486

(c) How many different sequences are possible if the second,third and fourth numbers must be the same?

6× 6× 1× 1× 6 = 216


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6× 6× 6× 6× 6 = 7776


6× 3× 3× 3× 3 = 486


6× 6× 1× 1× 6 = 216


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6× 6× 6× 6× 6 = 7776


6× 3× 3× 3× 3 = 486


6× 6× 1× 1× 6 = 216


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6× 6× 6× 6× 6 = 7776


6× 3× 3× 3× 3 = 486


6× 6× 1× 1× 6 = 216


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6× 6× 6× 6× 6 = 7776


6× 3× 3× 3× 3 = 486


6× 6× 1× 1× 6 = 216


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6× 6× 6× 6× 6 = 7776


6× 3× 3× 3× 3 = 486


6× 6× 1× 1× 6 = 216


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Math 1300: Section 7- 3 Basic Counting Principles