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Chapter 15

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

  2. - 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

  3. - 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

  4. - 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

  5. - 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??

  6. - Venn Diagram - U B S n(B) = n(S) = n(U) =

  7. - Venn Diagram - The intersection of the subsets are the people with brothers and sisters. U B S n(B) = n(S) =

  8. - Venn Diagram - The union of the subsets represents the people are in either set U S B Union -

  9. - 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:

  10. - 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:

  11. - Complement of the Union - The Confederacy! Haha

  12. - 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) =

  13. - Lets wrap it up - U B S n(B) = n(S) = n(U) =

  14. - 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?

  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

  16. - 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

  17. - 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

  18. - Practice -

  19. - Practice -

  20. - Practice -

  21. - 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)

  22. - Homework Questions -

  23. - 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

  24. - 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

  25. - Multiplication Principle - It is possible to select from the 3 shirts, 2 blazers and 5 pants by applying the multiplication principle: Outfits =

  26. - 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…

  27. - 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.

  28. - 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

  29. - 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

  30. - 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

  31. - 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

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

  33. - 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 ?

  34. - Homework Questions -

  35. - 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 .

  36. - 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

  37. - 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

  38. - 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?

  39. - 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

  40. - 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

  41. - 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?

  42. - 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

  43. - 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)

  44. - 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?

  45. - 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

  46. - Example - Consider MEXICO and CANADA Each of the letters in Mexico is different A is repeated three times in CANADA

  47. - Example - How many permutations are there for the letters of MASSACHUSETTS

  48. - 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

  49. - 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 =

  50. - 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)

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