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Chapter 15. Diagrams, Principals, Permutations, Combinations and Binomial Theorem Mr. Morrow 4/9/2013 – 4/19/2013. - Chapter Objectives -.

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Chapter 15

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Chapter 15

Chapter 15

Diagrams, Principals, Permutations, Combinations and Binomial Theorem

Mr. Morrow

4/9/2013 – 4/19/2013


Chapter 15

- Chapter Objectives -

To use Venn diagrams to illustrate intersections and unions of sets and to use the inclusion-exclusion principle to solve counting problems involving intersections and unions of sets.

To use the multiplication, addition and complement principles to solve counting problems.

To solve problems involving permutation and combinations

To solve counting problems that involve permutations with repetition and circular permutations

To use the binomial theorem and Pascal’s triangle


Chapter 15

- Combinatorics -

The theory of counting:

- Counting should be easy but it can get tricky when there are numerous objects to be counted

- Shortcuts (rules) make it easier to count large numbers of objects


Chapter 15

- Venn Diagrams -

These diagrams, in which you all should be familiar with, can be used to illustrate and separate sets of number we are counting

Definitions:

- The Universal Set is the set of ‘objects’ or elements to be counted and is often represented by the letter U

The number of elements in the universal set is represented by n(U)

In a Venn Diagram the universal set is represented as a rectangle

- A subsetis the group of elements that we are interested in

In a Venn Diagram each subset is represented by a circle


Chapter 15

- Class Poll -

Who has a brother or brothers?

Who has a sister or sisters?

Who has both brothers and sisters?

Who has neither brothers nor sisters?

We can represent this information using

a Venn Diagram

Who wishes they didn’t have their brothers or sisters??


Chapter 15

- Venn Diagram -

U

B

S

n(B) =

n(S) =

n(U) =


Chapter 15

- Venn Diagram -

The intersection of the subsets are

the people with brothers and sisters.

U

B

S

n(B) =

n(S) =


Chapter 15

- Venn Diagram -

The union of the subsets represents

the people are in either set

U

S

B

Union -


Chapter 15

- The Inclusion – Exclusion Principle -

U

S

B

Because the intersection is counted in each subset, when we determine the union we only count the intersection once therefore:


Chapter 15

- The Complement Principle -

U

S

B

The compliment of set B is written as either B’ or and is the elements not in set B:


Chapter 15

- Complement of the Union -

The Confederacy! Haha


Chapter 15

- Complement of the Union -

The complement of the subsets is the number of people who have neither brothers nor sisters

U

B

S

n(B) =

n(S) =

n(U) =


Chapter 15

- Lets wrap it up -

U

B

S

n(B) =

n(S) =

n(U) =


Chapter 15

- Example -

Of the 540 seniors at Central High School, 335 are taking math, 287 are taking science and 220 are taking both math and science. How many are taking neither math or science?


Chapter 15

- Example -

U

M

S

n(S) = 287

n(M) = 335

n(U) = 540

= 220

The number of students taking either math or science can be determined

using the inclusion – exclusion principle.

=335 + 287 – 220

= 402

To find the number of seniors taking neither math or science we find the

complement of the union which is = 540 – 402 = 138


Chapter 15

- Practice -

Given the set of number from 1 to 30. Determine the subsets of even numbers and numbers divisible by 3. Represent this information in a Venn Diagram.

- Determine

- Determine

- Determine

- Determine


Chapter 15

- Practice -

25

5

1

11

26

4

16

6

3

22

12

2

9

27

7

19

20

18

8

24

15

14

30

10

21

28

23

17

29

13


Chapter 15

- Practice -


Chapter 15

- Practice -


Chapter 15

- Practice -


Chapter 15

- Practice -

If A={3, 6, 9, 12, 15, 18}, B={2, 4, 6, 8, 10, 12, 14, 16, 18} and C={1, 4, 9, 16} list the elements in

{6, 12, 18}

{4,16}

{9}

{}

{1, 3, 4, 6, 9, 12, 15, 16, 18}

{2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18}

{4, 9, 16}

{1, 2, 4, 6, 8, 9, 10, 12, 14, 16, 18}

{2, 4, 6, 8, 9, 10, 12, 14, 16, 18}

{3, 4, 6, 9, 12, 15, 16, 18}

Homework: pg. 568 (# 1, 3, 11, 13, 15, 17, 19, 21)


Chapter 15

- Homework Questions -


Chapter 15

- Multiplication Principle -

The other day we saw how we can use Venn Diagrams to help us visualize concepts of counting. Right now we will see an idea called “Trees”

Example:

Try flipping a coin three times, how many possible outcomes are there?

H

There are eight possible outcomes:

HHH, HHT, HTH, HTT,

THH, THT, TTH, TTT

H

T

H

H

T

T

H

H

T

T

H

T


Chapter 15

- Multiplication Principle -

Definition: If an action can be performed in n1 ways and for each of these ways another action can be performed in n2 ways, then the two actions can be performed in n1n2 ways.

Practice:

How many of you girls stand in your

closet for at least 5 minutes a day?

This is widely not true, but assume

you only have three button-ups,

two blazers and five pairs of pants.

How many outfits are available?

Draw a tree to verify


Chapter 15

- Multiplication Principle -

It is possible to select from the 3 shirts, 2 blazers and 5 pants by applying the multiplication principle:

Outfits =


Chapter 15

- Practice -

The license plate math: In Virginia a standard (none custom) license plate is issued like the one below. How many combinations can be created?

Think about what you see before you answer…


Chapter 15

- Addition Principle -

We just saw how many possible license plates could be made using the picture below. Now consider that VA may decide to start making licenses such that the 1st 3 spaces are numbers and the last 4 spaces are letters. How many possible licenses would VA now have?

(3 letter/ 4 numbers)

(3 numbers/ 4 letters)

Ok…now what?

This example introduces the concept of mutually exclusive events. We can’t put both a letter and a number into each space – mutually exclusive events cannot be performed together.


Chapter 15

- Mutually Exclusive Venn Diagrams -

How do we represent mutually exclusive events?

The addition principle states that if two actions are mutually exclusive, and the first can be done n1ways and the second can be done n2 ways, the one action orthe other can be done in

n1+ n2 ways.

U

L

N


Chapter 15

- Addition Principle -

Soo…how many possible licenses would VA now have?

(3 letter/ 4 numbers)

(3 numbers/ 4 letters)

175,760,000 + 457,976,000 = 632,736,000


Chapter 15

- Factorials -

How many ways can I arrange the Harry Potter series books on a shelf?

… The first thing we need to know is how many books are in the series?

Ok, now think of having 7 slots on the shelf

How many books could go in

the first slot? second? third?

Why?

We multiply each by the

preceding term…

= 5040

7

6

5

4

3

2

1


Chapter 15

- Practice -

  • If 8! is 40,320 what is 9! How would we determine 23!

    • Go to the <run> menu on your calculator. Press the OPTN button. Press <F6> which is the arrow and then press <F3> which is PROB. Enter the factorial number then press <F1> to invoke the factorial calculation

  • If Costa Rica chose not to use 0 on its license plates, how many different plates are possible?

  • A guy has six ball caps and 3 hoodies

    • How many cap and hoodie combinations can he choose from?

    • If he chooses to wear a ball cap or a hoodie but not both, how many choices does he have?

  • If 10 runners compete in a race, in how many ways can 1st, 2nd and 3rd place prizes be awarded?

  • 9! = 362,880 and 23! = 2.58 x 1022

    Costa Rican tags = 9 x 9 x 9 x 9 x 9 = 59049

    Outfits = 6 x 3 = 18 or 6 + 3 = 9

    Prizes = 10 x 9 x 8 = 720


    Chapter 15

    - Homework -

    Page 575 – 577 (# 1, 3, 5, 7, 11, 13, 15, 17)


    Chapter 15

    - Warm Up-

    In an election-day survey of 100 voters leaving the polls, 52 said they voted for Proposition 1, and 38 said they voted for Proposition 2. If 18 said they voted for both, how many voted for neither?

    In a survey of 48 high school students, 20 liked classical music and 16 liked bluegrass music. Twenty students said they didn’t like either. How many liked classical but not bluegrass?

    For a universal set U, what is ?


    Chapter 15

    - Homework Questions -


    Chapter 15

    - Permutations-

    Given n symbols, how many non-repetitive lists of length rcan be made from the nsymbols in which order matters? By using the multiplication principle to obtain the answer

    By cancellation this value can also be written as:

    We summarize this as follows:

    The number of non-repetitive lists of length rwhose entries

    are chosen from a set of n possible entries is .


    Chapter 15

    - Permutations-

    For this class, you will represent this as:

    Where:

    n is the number of objects/ elements available to be chosen

    ris the number of objects/ elements actually chosen


    Chapter 15

    - Combinations -

    Without diving too deep into the math behind this next idea…

    Imagine we build from Permutations but instead we are curious about solutions in which order does not matter.

    For this class, you will represent this as:

    Where:

    nis the number of objects/ elements available to be chosen

    r is the number of objects/ elements actually chosen


    Chapter 15

    - Examples-

    A company advertises two job openings, one for a computer programmer and one for an IT specialist. If 10 people who are

    qualified for either position apply, in how many ways can the job openings be filled?

    A company advertises two job openings

    for computer programmers, both with

    the same salary and job description.

    In how many ways can the openings be

    filled if 10 people apply?


    Chapter 15

    - Examples-

    A company advertises two job openings, one for a computer programmer and one for an IT specialist. If 10 people who are

    qualified for either position apply, in how many ways can the job openings be filled?

    = 90

    A company advertises two job openings for computer programmers, both with the same salary and job description. In how many ways can the openings be filled if 10 people apply?

    = 45


    Chapter 15

    - Using a Calculator -

    • Your calculator will quickly compute permutations and combinations.

    • For example 1 we are attempting to calculate 10P2

      • Go to the Run menu

        • Press the OPTN button

        • Press F6 (arrow over to additional menu choices)

        • Press F3 (PROB)

        • Key in 10, press F2 (nPr) and key in 2

        • Press EXE

    • For example 2 we are attempting to calculate 10C2

      • Go to the Run menu

        • Press the OPTN button

        • Press F6 (arrow over to additional menu choices)

        • Press F3 (PROB)

        • Key in 10, press F3 (nCr) and key in 2

        • Press EXE


    Chapter 15

    - Practice-

    A single 5-card hand is dealt off of a standard 52-card deck. How many different 5-card hands are possible?

    Same scenario are (1), however, how many such hands are there in which two of the cards of clubs and three are hearts?

    Imagine a lottery that works as follows. A bucket contains 36 balls numbered 1, 2, 3, 4, …, 36. Six of these balls will be drawn randomly. For $1 you buy a ticket that has six blanks. You fill in the blanks with six different numbers between 1 and 36. You win $1,000,000 if you chose the same numbers that are drawn, regardless of the order. What are the changes of winning?


    Chapter 15

    - Practice-

    A single 5-card hand is dealt off of a standard 52-card deck. How many different 5-card hands are possible?

    = 2,598,960

    Same scenario are (1), however, how many such hands are there in which two of the cards of clubs and three are hearts?

    Think of such hand as being described by a list of length two of the form where the first entry is a 2-element subset of 13 club cards, and the second entry is a 3-element subset of the 13 heart cards.

    = 22,308


    Chapter 15

    - Practice-

    Imagine a lottery that works as follows. A bucket contains 36 calls numbered 1, 2, 3, 4, …, 36. Six of these balls will be drawn randomly. For $1 you buy a ticket that has six blanks. You fill in the blanks with six different numbers between 1 and 36. You win $1,000,000 if you chose the same numbers that are drawn, regardless of the order. What are the changes of winning?

    You are choosing six numbers from a set of 36 numbers

    = 1,947,792

    Homework: pg. 580 (# 1-19 odd)


    Chapter 15

    - Ummm?? -

    • Write down all the possible arrangements of the letters MOP

      • How many arrangements were you able to create?

    • Write down all the distinguishable arrangements of the letters MOM

      • How many arrangements were you able to create?

    • Why was the total number of arrangements

    • different for the two sets?


    Chapter 15

    - Permutations with Repetition-

    If an element is repeated in an arrangement, then fewer permutations result.

    Where:

    n1= the number of elements of type 1,

    n2= the number of elements of type 2 …

    nk= the number of elements of type k.

    This gives us the number of distinguishable permutations of n elements


    Chapter 15

    - Example -

    Consider MEXICO and CANADA

    Each of the letters in Mexico is different

    A is repeated three times in CANADA


    Chapter 15

    - Example -

    How many permutations are there for the letters of

    MASSACHUSETTS


    Chapter 15

    - Practice-

    A person is at point X on the grid below and is going to walk to point Y by always traveling south or east. How many routes from X to Y are possible?

    X

    Y


    Chapter 15

    - Wait a minute? -

    How many circular permutations are possible when seating four people around a table?

    B

    A

    D

    C

    A

    C

    D

    B

    C

    A

    B

    D

    C

    B

    A

    D

    DABC

    ABCD

    CDAB

    BCDA

    • Circular permutations are the same because A

    • is always to the right of B which is to the right of C which

    • is to the right of D.

    • Linear permutations are different

    • Therefore the number of circular permutations =


    Chapter 15

    - Practice-

    How many different ways can five children arrange themselves for a game of ring-around-the-rosie?

    How many ways can ten people

    be seated around a circular table

    if the host and hostess can’t be

    seated together?

    Homework pg. 585 (#1-15 odd)


    Chapter 15

    - Practice-

    10

    8

    7

    7

    1

    2

    6

    3

    5

    4


    Chapter 15

    - Warm Up -


    Chapter 15

    - Homework Questions-


    Chapter 15

    - Investigation -

    Determine:

    Can you see any patterns?

    How do the number of terms relate to the exponent?

    Is there a pattern to the exponents in each term?

    Do you detect any relationship for the coefficients of each term?


    Chapter 15

    - Aint nobody got time to do all that-

    Our objective now is to develop a method for expanding binomials of the where n is a positive integer

    When we expand (FOIL) the linear factors in a binomial the result is a polynomial with n+1 terms

    Each term in the polynomial is of the form where k is a coefficient and y and z are exponents.

    The hard part is finding a simple way to determine k, y, and z.


    Chapter 15

    - Binomial Theorem-

    If n is a positive integer then:

    Where each term is in the form:

    - Therefore for any term t of a binomial to the power the coefficient

    - The sum of the exponents of a and b is always n. (y + z = n)

    - The exponent of the first monomial of ‘a’, y = n – (t – 1)

    - The exponent of the second monomial of ‘b’, z = t – 1


    Chapter 15

    - Example -

    Find the first four terms in the expansion of

    1st term

    k = 21C(1-1) = 21C0 = 1

    y = 21 – (1 – 1) = 21

    z = 1 – 1 = 0

    therefore the first term is the product of 1 x a21 x b0 = a21

    2nd term

    k = 21C(2-1) = 21C1 = 21

    y = 21 – (2 – 1) = 20

    z = 2 – 1 = 1

    therefore the second term is 21a20b1 = 21a20b

    3rd term

    k = 21C(3 – 1) = 21C2 = 210

    y = 21 – (3 – 1) = 19

    z = 3 – 1 = 2

    therefore the third term is 210a19b2

    4th term

    k = 21C(4 – 1) = 21C3 = 1330

    y = 21 – (4 – 1) = 18

    z = 4 – 1 = 3

    therefore the fourth term is 1330a18b3


    Chapter 15

    - Example -

    • What is the 8th term of ?

      • k = 10C8-1

      • k = 10C7

      • k = 120

      • The exponent for x = 10 – (8 – 1) = 3

      • The exponent for -2y = 8 – 1 = 7

      • Therefore the 8th term is

      • = 120x3(-2y)7

      • = 120x3(-2)7y7

      • = 120(-128)x3y7

      • = - 15360x3y7


    Chapter 15

    - Practice-

    • Give the expansion of (2a – b2)8

    • In the expansion of (a + b)20, what is the coefficient ofa) a17b3b) a3b17

    • Find the first four terms of the expansion of the expression (sinx – cosy)30. Do not simplify.

    • Find the value of (0.99)5 to the nearest hundredth by considering the expansion of (1 – 0.01)5


    Chapter 15

    - Practice-

    Expanding (2a – b2)8 we know that there are 8 + 1 = 9 terms.

    1st term = 8C0(2a)8(-b2)0 = 256a8

    2nd term = 8C1(2a)7(-b2)1 = (8)(-128)a7b2 = -1024a7b2

    3rd term = 8C2(2a)6(-b2)2 = (28)(64)a6b4 = 1792a6b4

    4th term = 8C3(2a)5(-b2)3 = (56)(-32)a5b6 = -1792a5b6

    5th term = 8C4(2a)4(-b2)4 = (70)(16)a4b8 = 1120a4b8

    6th term = 8C5(2a)3(-b2)5 = (56)(-8)a3b10 = -448a3b10

    7th term = 8C6(2a)2(-b2)6 = (28)(4)a2b12 = 112a2b12

    8th term = 8C7(2a)1(-b2)7 = (8)(-2)ab14 = -16ab14

    9th term = 8C8(2a)0(-b2)8 = (1)(1)b16 = b16

    (2a – b2)8 = 256a8 - 1024a7b2 +1792a6b4 -1792a5b6 +1120a4b8 -448a3b10 +112a2b12 -16ab14 +b16


    Chapter 15

    - Practice-

    In the expansion of (a + b)20, what is the coefficient of

    a17b3 b) a3b17

    a) y = n – (t – 1)

    17 = 20 – (t – 1)

    t = 4

    k = nCt-1

    k = 20C3

    k = 1140

    b) y = n – (t – 1)

    3 = 20 – (t – 1)

    t = 18

    k = nCt-1

    k = 20C17

    k = 1140


    Chapter 15

    - Practice-

    Find the first four terms of the expansion of the expression(sinx– cosy)30. Do not simplify.

    1stterm = 30C0(sinx)30(-cosx)0 2ndterm = 30C1(sinx)29(-cosx)1

    3rd term = 30C2(sinx)28(-cosx)2 4thterm = 30C3(sinx)27(-cosx)3

    1st Term

    k = 30C0 = 1

    a = 30 – (1 – 1) = 30

    b = 1 – 1 = 0

    kxaby = sin30x

    2nd Term

    k = 30C1 = 30

    a = 30 – (2 – 1) = 29

    b = 2 – 1 = 1

    kxaby = -30sin29xcosy

    3rdTerm

    k = 30C2= 435

    a = 30 – (3 – 1) = 28

    b = 3 – 1 = 2

    kxaby = 435sin28xcos2y

    4th Term

    k = 30C3= 4060

    a = 30 – (4 – 1) = 27

    b = 4 – 1 = 3

    kxaby = -4060sin27xcos3y


    Chapter 15

    - Practice-

    Find the value of (0.99)5 to the nearest hundredth by considering the expansion of (1 – 0.01)5

    (1 – 0.01)5 = 5C0(1)5(-.01)0 + 5C1(1)4(-.01)1 + 5C2(1)3(-.01)2 + 5C3(1)2(-.01)3 + 5C4(1)1(-.01)4 + 5C5(1)0(-.01)5

    = 1 - .05 + .001 – .00001 + .00000005 - .0000000001

    = 0.9509900499

    Homework: pg. 592 (#1-21 odd)


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