1. Jeff Bivin -- LZHS Permutation, Combinations, Probability, Oh My… Jeff Bivin Lake Zurich High School jeff.bivin@lz95.org ICTM Conference October 21, 2011

2. Jeff Bivin -- LZHS

3. Fundamental Counting Principal How many different meals can be made if 2 main courses, 3 vegetables, and 2 desserts are available? M 1 M 2 V 1 V 2 V 3 D 1 D 2 D 1 D 2 D 1 D 2 1 2 3 4 5 6 7 8 9 10 11 12 Let’s choose a main course Now choose a vegetable Finally choose A dessert

4. Jeff Bivin -- LZHS Linear Permutations A club has 30 members and must select a president, vice president, secretary, and treasurer. How many different sets of officers are possible? presidentvice-presidentsecretarytreasurer

5. Jeff Bivin -- LZHS A club has 30 members and must select a president, vice president, secretary, and treasurer. How many different sets of officers are possible? presidentvice-presidentsecretarytreasurer 30 P 4 Linear Permutations

6. Jeff Bivin -- LZHS Permutation Formula

7. Jeff Bivin -- LZHS Linear Permutations There are 25 students in a classroom with 25 seats in the room, how many different seating charts are possible? seat 1seat 2seat 3seat 4seat 5 1.5511 x 10 25

8. Jeff Bivin -- LZHS Linear Permutations There are 25 students in a classroom with 25 seats in the room, how many different seating charts are possible? seat 1seat 2seat 3seat 4seat 5 25 P 25 1.5511 x 10 25

9. Jeff Bivin -- LZHS Permutation Formula 1.5511 x 10 25

10. Jeff Bivin -- LZHS Circular Permutations There are 5 people sitting at a round table, how many different seating arrangements are possible? straight line Divide by 5

11. Jeff Bivin -- LZHS Circular Permutations There are 5 people sitting at a round table, how many different seating arrangements are possible? straight line Treat all permutations as if linear Now consider the circular issue When circular, divide by the number of items in the circle

12. Jeff Bivin -- LZHS Circular Permutations There are 9 people sitting around a campfire, how many different seating arrangements are possible? straight line Treat all permutations as if linear Is it circular? Yes, divide by 9

13. Jeff Bivin -- LZHS There are 5 people sitting at a round table with a captain chair, how many different seating arrangements are possible? More Permutations straight line NOTE:

14. Jeff Bivin -- LZHS More Permutations How many ways can you arrange 3 keys on a key ring? straight line Treat all permutations as if linear Is it circular? Now, try it... PROBLEM: Turning it over results in the same outcome. Yes, divide by 3 So, we must divide by 2.

15. Jeff Bivin -- LZHS More Permutations How many ways can you arrange the letters MATH ? How many ways can you arrange the letters ABCDEF ?

16. Jeff Bivin -- LZHS Permutations with Repetition How many ways can you arrange the letters AAAB? Let’s look at the possibilities: AAAB AABA ABAA BAAA Are there any others? What is the problem? If a permutation has repeated items, we divide by the number of ways of arranging the repeated items (as if they were different). Divide by 3!

17. Jeff Bivin -- LZHS How many ways can you arrange 5 red, 7 blue and 8 white flags on the tack strip across the front of the classroom? If all were different, how may ways could we arrange 20 items? There are 5 repeated red flags  Divide by 5! There are 7 repeated blue flags  Divide by 7! There are 8 repeated white flags  Divide by 8!

18. Jeff Bivin -- LZHS How many ways can you arrange the letters in the non-word A A B B C C C C D E F G G G G G G ? If all were different, how may ways could we arrange 17 items? There are 2 repeated A’s  Divide by 2! There are 2 repeated B’s  Divide by 2! There are 4 repeated C’s  Divide by 4! There are 6 repeated G’s  Divide by 6!

19. Jeff Bivin -- LZHS Permutations ORDER Multiply the possibilities Divide by the number of items in the circle Divide by 2 Divide by the factorial of the number of each duplicated item Assume the items are in a straight line ! Use the n P r formula (if no replacement) or Are the items in a circle? ? Can the item be turned over? ? Are there duplicate items in your arrangement? ?

20. Jeff Bivin -- LZHS How many ways can you put 5 red and 7 brown beads on a necklace? How may ways could we arrange 12 items in a straight line? Is it circular? Yes  divide by 12 Can it be turned over? Yes  divide by 2 Are there repeated items? Yes  divide by 5! and 7! 33

21. Jeff Bivin -- LZHS How many ways can you arrange 5 red and 7 brown beads on a necklace that has a clasp? How may ways could we arrange 12 items in a straight line? Is it circular? N0  the clasp makes it linear Can it be turned over? Yes  divide by 2 Are there repeated items? Yes  divide by 5! and 7! 396

22. Jeff Bivin -- LZHS How different license plates can have 2 letters followed by 3 digits (no repeats)? A straight line? Is it circular? No Can it be turned over? No Are there repeated items? No 468,000 26∙25∙10∙9∙8 letter number

23. Jeff Bivin -- LZHS How different license plates can have 2 letters followed by 3 digits with repeats? A straight line? Is it circular? No Can it be turned over? No Are there repeated items? Yes, but because we are using multiplication and not factorials, we do not need to divide by anything. 676,000 26∙ ∙10∙ ∙ letter number

24. Jeff Bivin -- LZHS

25. Combinations NO order NO replacement Use the n C r formula Typically

26. Jeff Bivin -- LZHS Combinations An organization has 30 members and must select a committee of 4 people to plan an upcoming function. How many different committees are possible?

27. Jeff Bivin -- LZHS Combinations A plane contains 12 points, no three of which are co- linear. How many different triangles can be formed?

28. Jeff Bivin -- LZHS An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 red marbles? Combinations 5 red 6 white 9 blue 3 red havewant

29. Jeff Bivin -- LZHS An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 blue marbles? Combinations 5 red 6 white 9 blue 3 blue havewant

30. Jeff Bivin -- LZHS The OR factor. 5 red 6 white 9 blue 3 red OR 3 blue havewant OR  ADD An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 3 red marbles or 3 blue marbles?

31. Jeff Bivin -- LZHS The OR factor. 5 red 8 blue 4 red OR 4 blue have OR  ADD want An jar contains 13 marbles – 5 red and 8 blue. If four are selected at random, how many ways can you select 4 red marbles or 4 blue marbles?

32. Jeff Bivin -- LZHS An jar contains 20 marbles – 5 red, 6 white and 9 blue. If three are selected at random, how many ways can you select 1 red marble and 2 blue marbles? The AND factor. AND  MULTIPLY 5 red 6 white 9 blue 1 red and 2 blue havewant

33. Jeff Bivin -- LZHS 3 B 2 NB or 4 B 1 NB or 5 B At least 3 or 4 or 5 blue An jar contains 20 marbles – 5 red, 6 white and 9 blue. If five marbles are selected at random, how many ways can you select at least 3 blue marbles?

34. Jeff Bivin -- LZHS 0 R 5 Nr or 1 R 4 NR At most 0 or 1 red An jar contains 20 marbles – 5 red, 6 white and 9 blue. If five marbles are selected at random, how many ways can you select at most 1 red marbles?

35. Jeff Bivin -- LZHS And More

36. Jeff Bivin -- LZHS (x + y) 0 = 1 (x + y) 1 = 1x + 1y (x + y) 2 = 1x 2 + 2xy + 1y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3xy 2 +1y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4 (x + y) 5 = 1x 5 + 5x 4 y + 10x 3 y 2 + 10x 2 y 3 + 5xy 4 + 1y 5 (x + y) 6 = 1x 6 + 6x 5 y + 15x 4 y 2 + 20x 3 y 3 + 15x 2 y 4 + 6xy 5 + 1y 6 Let’s look at (x + y) p (x + y) 7 = _x 7 + _x 6 y + _x 5 y 2 + _x 4 y 3 + _x 3 y 4 + _x 2 y 5 + _xy 6 + _y 7

37. Jeff Bivin -- LZHS (x + y) 0 = 1 (x + y) 1 = 1x + 1y (x + y) 2 = 1x 2 + 2xy + 1y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3xy 2 +1y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4 (x + y) 5 = 1x 5 + 5x 4 y + 10x 3 y 2 + 10x 2 y 3 + 5xy 4 + 1y 5 (x + y) 6 = 1x 6 + 6x 5 y + 15x 4 y 2 + 20x 3 y 3 + 15x 2 y 4 + 6xy 5 + 1y 6 Let’s look at (x + y) p

38. Jeff Bivin -- LZHS (x + y) 0 = 1 (x + y) 1 = 1x + 1y (x + y) 2 = 1x 2 + 2xy + 1y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3xy 2 +1y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4 (x + y) 5 = 1x 5 + 5x 4 y + 10x 3 y 2 + 10x 2 y 3 + 5xy 4 + 1y 5 (x + y) 6 = 1x 6 + 6x 5 y + 15x 4 y 2 + 20x 3 y 3 + 15x 2 y 4 + 6xy 5 + 1y 6 Let’s look at (x + y) p

39. Jeff Bivin -- LZHS (x + y) 7 = 1 7 21 35 35 21 7 1 (x + y) 0 = 1 (x + y) 1 = 1 1 (x + y) 2 = 1 2 1 (x + y) 3 = 1 3 3 1 (x + y) 4 = 1 4 6 4 1 (x + y) 5 = 1 5 10 10 5 1 (x + y) 6 = 1 6 15 20 15 6 1 (x + y) 8 = 1 8 28 56 70 56 28 8 1 Let’s look at (x + y) p

40. Jeff Bivin -- LZHS (x + y) 7 = 1x 7 + 7x 6 y + 21x 5 y 2 + 35x 4 y 3 + 35x 3 y 4 + 21x 2 y 5 + 7xy 6 + 1y 7 (x + y) 0 = 1 (x + y) 1 = 1x + 1y (x + y) 2 = 1x 2 + 2xy + 1y 2 (x + y) 3 = 1x 3 + 3x 2 y + 3xy 2 +1y 3 (x + y) 4 = 1x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + 1y 4 (x + y) 5 = 1x 5 + 5x 4 y + 10x 3 y 2 + 10x 2 y 3 + 5xy 4 + 1y 5 (x + y) 6 = 1x 6 + 6x 5 y + 15x 4 y 2 + 20x 3 y 3 + 15x 2 y 4 + 6xy 5 + 1y 6 Let’s look at (x + y) p

41. Jeff Bivin -- LZHS In how many ways can you arrange the letters in the word M A T H E M A T I C A L ?

42. Jeff Bivin -- LZHS In how many ways can you arrange the letters in the non-word xxxxyyy? In how many ways can you arrange the letters in the non-word xxyyyyy? (x + y) 7 = 1x 7 + 7x 6 y + 21x 5 y 2 + 35x 4 y 3 + 35x 3 y 4 + 21x 2 y 5 + 7xy 6 + y 7

43. Jeff Bivin -- LZHS Let’s look closer (x + y) 7 = 1x 7 + 7x 6 y + 21x 5 y 2 + 35x 4 y 3 + 35x 3 y 4 + 21x 2 y 5 + 7xy 6 + y 7

44. Jeff Bivin -- LZHS An alternate look (x + y) 7 = 1x 7 + 7x 6 y + 21x 5 y 2 + 35x 4 y 3 + 35x 3 y 4 + 21x 2 y 5 + 7xy 6 + y 7

45. Jeff Bivin -- LZHS (2x - y) 4 = 16x 4 - 32x 3 y + 24x 2 y 2 - 8xy 3 + y 4

46. Jeff Bivin -- LZHS (3x + 2y) 5 = 243x 5 + 810x 4 y + 1080x 3 y 2 + 720x 2 y 3 + 240xy 4 + 32y 5

47. Jeff Bivin -- LZHS In how many ways can you arrange the letters in the non-word x x x x x y y y y y y y y y y ? Given: (x + y) 15 What is the coefficient of the term ____ x 5 y 10 ?

48. Jeff Bivin -- LZHS Given: (4x - 3y) 10 In how many ways can you arrange the letters in the non-word x x x x x x x y y y ? What is the coefficient of the term ____ x 7 y 3 ?

49. Jeff Bivin -- LZHS Expand: (x + y + z) 2 (x + y + z) x 2 + xy + xz + yx + y 2 + yz + zx + zy + z 2 x 2 + 2xy + 2xz + y 2 + 2yz + z 2

50. Jeff Bivin -- LZHS x 3 + 3x 2 y + 3x 2 z + 3xy 2 + 6xyz + 3xz 2 + y 3 + 3y 2 z + 3yz 2 + z 3 (x + y + z) 2 (x + y + z) x 3 + x 2 y + x 2 z + 2x 2 y + 2xy 2 + 2xyz + 2x 2 z + 2xzy + 2xz 2 + y 2 x + y 3 + y 2 z +2yzx + 2y 2 z + 2yz 2 + z 2 x + z 2 y + z 3 Expand: (x + y + z) 3 (x 2 + 2xy + 2xz + y 2 + 2yz + z 2 )(x + y + z) We did this in the last example

51. Jeff Bivin -- LZHS x 3 + 3x 2 y + 3x 2 z + 3xy 2 + 6xyz + 3xz 2 + y 3 + 3y 2 z + 3yz 2 + z 3 Given: (x + y + z) 3 In how many ways can you arrange the letters in the non-word xyz ? What is the coefficient of the term ____ xyz ? In how many ways can you arrange the letters in the non-word xxz ? What is the coefficient of the term ____ x 2 z ? REMEMBER

52. Jeff Bivin -- LZHS Given: (x + y + z) 15 In how many ways can you arrange the letters in the non-word xxyyyyyyyzzzzzz ? What is the coefficient of the term ____ x 2 y 7 z 6 ?

53. Jeff Bivin -- LZHS Given: (2x + 3y - z) 9 In how many ways can you arrange the letters in the non-word x x x y y y y z z ? What is the coefficient of the term ____ x 3 y 4 z 2 ?

54. Jeff Bivin -- LZHS In how many ways can you arrange the letters in the non-word a a a a a b b b b b b c c c c c c c d d ? Given: (a + b + c + d) 20 What is the coefficient of the term ____ a 5 b 6 c 7 d 2 ?

55. Jeff Bivin -- LZHS

56. Can we think of combinations as permutations with repetitions? In a group of 7 people, how many different committees of 3 people can we select? Two choices…. A member of the 3 person committee A member of the 4 person non-committee

57. Jeff Bivin -- LZHS In a group of 7 people, how many different committees of 3 people can we select? AlexBettyChuckDebEdFionaGabe NCNCNNC How many ways can you arrange 3 C’s and 4 N’s?

58. Jeff Bivin -- LZHS In a group of 7 people, a pair of 2 people are needed for one committee and 3 different people are need for a second committee. How many possibilities exist? AlexBettyChuckDebEdFionaGabe C1NC2C1NC2N How many ways can you arrange 2 C1’s and 2 C’s and 3 N’s?

59. Jeff Bivin -- LZHS In a group of 7 people, how may ways can a president, a vice-president and a secretary be selected? AlexBettyChuckDebEdFionaGabe NVPNPNNS How many ways can you arrange 1 P and 1 VP and 1 S and 4 N’s? Can we think of most permutations as permutations with repetitions?

60. Jeff Bivin -- LZHS

61. PROBABILITY Defined number of success total number of outcomes The ratio 

62. Jeff Bivin -- LZHS Probability A coin is tossed, what is the probability that you will obtain a heads? Look at the sample space/possible outcomes: { H, T } number of success total number of outcomes Pr(H) = 1 2 =

63. Jeff Bivin -- LZHS number of success Probability A die is tossed, what is the probability that you will obtain a number greater than 4? Look at the sample space/possible outcomes: total number of outcomes Pr(>4) = 2 6 = 1 3 = { 1, 2, 3, 4, 5, 6 }

64. Jeff Bivin -- LZHS number of failures total number of outcomes number of success Probability – Success & Failure A die is tossed, what is the probability that you will obtain a number greater than 4? Pr(>4) = 2 6 = 1 3 = What is the probability that you fail to obtain a number greater than 4? Pr(>4) = 4 6 2 3 = = TOTAL = Pr(success) + Pr(failure) = 1

65. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are red? number of success total number of outcomes Probability Pr(3 R ) = 5C35C3 13 C 3 = 5 red 8 blue havewant 3 red Total: 13  3

66. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are blue? number of success total number of outcomes Probability Pr(3 B ) = 8C38C3 13 C 3 = 5 red 8 blue havewant 3 blue Total: 13  3

67. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one is red and two are blue? number of success total number of outcomes Probability – “and” Pr(1 R 2 B ) = 5 C 1 ● 8 C 2 13 C 3 = 5 red 8 blue havewant 1 red Total: 13  3 2 blue multiply

68. Jeff Bivin -- LZHS A jar contains 5 red, 8 blue and 7 white marbles. If 3 marbles are selected at random, what is the probability that one of each color is selected? # of success total # of outcomes Pr(1 R,1 B,1 W ) = 5C1●8C1●7C15C1●8C1●7C1 20 C 3 = 5 red 8 blue 7 white havewant 1 red Total: 20  3 1 blue 1 white 1 red, 1 blue, & 1 white

69. Jeff Bivin -- LZHS A jar contains 7 red, 5 blue and 3 white marbles. If 4 marbles are selected at random, what is the probability that 2 red and 2 white marbles are selected? # of success total # of outcomes Pr(2 R,2 W ) = 7 C 2 ● 3 C 2 15 C 4 = 7 red 5 blue 3 white havewant 2 red Total: 15  4 2 white

70. Jeff Bivin -- LZHS Five cards are dealt from a standard deck of cards. What is the probability that 3 hearts and 2 clubs are obtained? # of success total # of outcomes Pr(3 H,2 C ) = 13 C 3 ● 13 C 2 52 C 5 = 13 diamonds 13 hearts 13 clubs 13 spades havewant 3 hearts Total: 52  5 2 clubs

71. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that all three are red or all three are blue? # of success total # of outcomes Probability – “or” Pr(3 R or 3 B ) = 5 C 3 + 8 C 3 13 C 3 = 5 red 8 blue havewant 3 red Total: 13  3 3 blue want OR

72. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles and 7 yellow marbles. If 3 marbles are selected at random, what is the probability that all three are the same color? 5 C 3 + 8 C 3 + 7 C 3 # of success total # of outcomes Pr(3 R or 3 B or 3 w ) = 20 C 3 = 5 red 8 blue 7 yellow havewant 3 red Total: 20  3 3 blue want OR 3 red or 3 blue or 3 yellow ? want 3 yellow OR

73. Jeff Bivin -- LZHS 26 C 2 + 4 C 2 – 2 C 2 # of success total # of outcomes Probability – “or” with overlap Pr(2 R or 2 B ) = Pr(2 R ) + Pr(2 K ) – Pr(2 RK ) 52 C 2 = 26 red 26 black havewant 2 red Total: 52  2 2 kings want OR 2 red kings overlap If two cards are selected from a standard deck of cards, what is the probability that both are red or both are kings? 4 kings 48 other

74. Jeff Bivin -- LZHS 5 C 2 ● 8 C 1 + 5 C 1 ● 8 C 2 # of success total # of outcomes Probability – “and” with “or” Pr(2 R 1 B or 1 R 2 B ) = 13 C 3 = 5 red 8 blue havewant 2 red Total: 13  3 1 blue want OR 1 red 2 blue A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that two are red and one is blue or that one is red and two are blue?

75. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least two red marbles are selected? 5 C 2 ● 8 C 1 + 5 C 3 # of success total # of outcomes Probability – “at least” Pr(at least 2 Red ) = 13 C 3 = 5 red 8 blue havewant 2 red Total: 13  3 1 blue want OR 3 red 2 red or 3 red 2 red and 1 blue or 3 red Pr(2 R 1 B or 3 R ) =

76. Jeff Bivin -- LZHS 5 C 1 ● 8 C 2 + 5 C 2 ● 8 C 1 + 5 C 3 A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least one red marble is selected? Probability – “at least” Pr(at least 1 Red ) = 13 C 3 5 red 8 blue havewant 1 red Total: 13  3 2 blue want OR 2 red Pr(1 R 2 B or 2 R 1 B or 3 R ) = want OR 3 red 1 blue

77. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that NO red marbles are selected? 8C38C3 Probability – “at least” Pr(0 R 3 B ) = 13 C 3 5 red 8 blue havewant Total: 13  3 3 blue In the previous example we found Pr(success) + Pr(failure) = 1

78. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that at least one red marble is selected? Probability – “at least” Pr(>1 red) = 1 – Pr( 0 red ) Pr(success) + Pr(failure) = 1 Pr(success) = 1 - Pr(failure) Pr(3 blue)

79. Jeff Bivin -- LZHS A jar contains 8 red and 9 blue marbles. If 7 marbles are selected at random, what is the probability that at least one red marbles is selected? Probability – “at least” Pr(at least 1 Red ) Pr(1 R 6 B or 2 R 5 B or 3 R 4 B or 4 R 3 B or 5 R 2 B or 6 R 1 B or 7 R ) Pr(0 Red ) Pr( 0 R 7 B ) success failure FASTEST Pr(at least 1 Red ) = 1 - Pr(0 R 7 B ) =

80. Jeff Bivin -- LZHS A jar contains 8 red, 9 blue and 3 white marbles. If 7 marbles are selected at random, what is the probability that at least three red marbles are selected? Probability – “at least” Pr(> 3 Red )  Pr(3-7 red) Pr(< 3 Red )  Pr(0-2 red) success failure FASTEST 1 - Pr(0 R 7 NR or 1 R 6 NR or 2 R 5 NR )

81. Jeff Bivin -- LZHS Probability – “with replacement” Must use fractions! R B B Note: In this example an order is specified A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one red followed by two blue marbles are selected if each marble is replaced after each selection?

82. Jeff Bivin -- LZHS A jar contains 5 red and 8 blue marbles. If 3 marbles are selected at random, what is the probability that one red and two blue marbles are selected if each marble is replaced after each selection? Probability – “with replacement” Must use fractions! Must account of any order! Problem: Fractions imply order! R B B