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Permutation, Combinations, Probability, Oh My…

Permutation, Combinations, Probability, Oh My…. Jeff Bivin Lake Zurich High School jeff.bivin@lz95.org ICTM Conference October 21, 2011. Permutations. Fundamental Counting Principal.

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Permutation, Combinations, Probability, Oh My…

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

  2. Permutations

  3. Fundamental Counting Principal How many different meals can be made if 2 main courses, 3 vegetables, and 2 desserts are available? Let’s choose a main course 2 M1 M2 Now choose a vegetable 3 x V1 V2 V3V1 V2 V3 Finally choose A dessert 2 x D1 D2 D1 D2 D1 D2 D1 D2 D1 D2 D1 D2 12 1 2 3 4 5 6 7 8 9 10 11 12

  4. 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? 30 29 28 27 * * * president vice-president secretary treasurer 657,720

  5. Linear Permutations Alternative A club has 30 members and must select a president, vice president, secretary, and treasurer. How many different sets of officers are possible? 1 2 3 4 30 29 28 27 * * * Try with your calculator president vice-president secretary treasurer 657,720 30P4

  6. Permutation Formula 657,720

  7. Linear Permutations There are 25 students in a classroom with 25 seats in the room, how many different seating charts are possible? 25 24 23 22 21 * * * * . . . seat 1 seat 2 seat 3 seat 4 seat 5 1.5511 x 1025 25!

  8. Linear Permutations Alternative There are 25 students in a classroom with 25 seats in the room, how many different seating charts are possible? 25 24 23 22 21 * * * * . . . seat 1 seat 2 seat 3 seat 4 seat 5 1.5511 x 1025 25! 25P25

  9. Permutation Formula 1.5511 x 1025

  10. Circular Permutations There are 5 people sitting at a round table, how many different seating arrangements are possible? straight line Divide by 5 A B C D E E A B C D D E A B C C D E A B B C D E A

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

  12. Circular Permutations There are 9 people sitting around a campfire, how many different seating arrangements are possible? straight line Yes, divide by 9 Treat all permutations as if linear Is it circular? A B C D E F G H I

  13. More Permutations There are 5 people sitting at a round table with a captain chair, how many different seating arrangements are possible? straight line NOT CIRCULAR A B C D E E A B C D D E A B C NOTE: C D E A B B C D E A Each table has someone different in the captian chair!

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

  15. More Permutations How many ways can you arrange the letters MATH ? How many ways can you arrange the letters ABCDEF ?

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

  17. 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. 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. ! Assume the items are in a straight line ? Are the items in a circle? ? Can the item be turned over? ? Are there duplicate items in your arrangement? Permutations ORDER Multiply the possibilities or Use the nPrformula (if no replacement) Divide by the number of items in the circle Divide by 2 Divide by the factorial of the number of each duplicated item

  20. 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 33 Are there repeated items? Yes  divide by 5! and 7!

  21. 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 2 396 Are there repeated items? Yes  divide by 5! and 7!

  22. How different license plates can have 2 letters followed by 3 digits (no repeats)? A straight line? Is it circular? No 468,000 Can it be turned over? No Are there repeated items? No

  23. How different license plates can have 2 letters followed by 3 digits with repeats? A straight line? Is it circular? No 676,000 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.

  24. Combinations

  25. Combinations NO order NO replacement Use the nCrformula Typically

  26. 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? 27,405

  27. Combinations A plane contains 12 points, no three of which are co-linear. How many different triangles can be formed? 220

  28. Combinations 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? have want 5 red 6 white 9 blue 3 red 10

  29. Combinations 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? have want 5 red 6 white 9 blue 84 3 blue

  30. The OR factor. 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? OR ADD have want want 5 red 6 white 9 blue 3 red OR 3 blue

  31. The OR factor. 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? OR ADD want have want 5 red 8 blue 4 red OR 4 blue

  32. The AND factor. 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? AND MULTIPLY have want 5 red 6 white 9 blue 1 red and 2 blue

  33. At least 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? 3 or 4 or 5 blue 3B2NB or 4B1NB or 5B

  34. At most 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? 0 or 1 red 0R5Nror 1R4NR

  35. BinomialExpansion And More

  36. Let’s look at (x + y)p (x + y)0 = 1 Look at the exponents! (x + y)1 = 1x + 1y (x + y)2 = 1x2 + 2xy + 1y2 (x + y)3 = 1x3 + 3x2y + 3xy2 +1y3 (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 (x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6 (x + y)7 = _x7 + _x6y + _x5y2 + _x4y3 + _x3y4 + _x2y5 + _xy6 + _y7

  37. Let’s look at (x + y)p (x + y)0 = 1 Look at the coefficients! (x + y)1 = 1x + 1y (x + y)2 = 1x2 + 2xy + 1y2 (x + y)3 = 1x3 + 3x2y + 3xy2 +1y3 (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 (x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6

  38. Let’s look at (x + y)p (x + y)0 = 1 Look at the coefficients! (x + y)1 = 1x + 1y (x + y)2 = 1x2 + 2xy + 1y2 (x + y)3 = 1x3 + 3x2y + 3xy2 +1y3 (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 (x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6

  39. Let’s look at (x + y)p (x + y)0 = 1 Look at the coefficients! (x + y)1 = 1 1 (x + y)2 = 12 1 (x + y)3 = 1331 PASCAL'S TRIANGLE (x + y)4 = 14 641 (x + y)5 = 15 101051 (x + y)6 = 161520156 1 (x + y)7 = 1721353521 7 1 (x + y)8 = 1828567056 28 8 1

  40. Let’s look at (x + y)p (x + y)0 = 1 Let's Apply Pascal's Triangle (x + y)1 = 1x + 1y (x + y)2 = 1x2 + 2xy + 1y2 (x + y)3 = 1x3 + 3x2y + 3xy2 +1y3 (x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 (x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5 (x + y)6 = 1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6 (x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + 1y7

  41. Remember Permutations with repetitions 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. 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 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

  43. Let’s look closer (x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

  44. An alternate look (x + y)7 = 1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + y7

  45. (2x - y)4 = 16x4 - 32x3y + 24x2y2 - 8xy3 + y4

  46. (3x + 2y)5 = 243x5 + 810x4y + 1080x3y2 + 720x2y3 + 240xy4 + 32y5

  47. Given: (x + y)15 What is the coefficient of the term ____ x5y10 ? 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 ?

  48. Given: (4x - 3y)10 What is the coefficient of the term ____ x7y3 ? In how many ways can you arrange the letters in the non-word x x x x x x x y y y ?

  49. Expand: (x + y + z)2 (x + y + z) (x + y + z) x2 + xy + xz + yx + y2 + yz + zx + zy + z2 x2 + 2xy + 2xz + y2 + 2yz + z2

  50. We did this in the last example Expand: (x + y + z)3 (x + y + z)2 (x + y + z) (x2 + 2xy + 2xz + y2 + 2yz + z2)(x + y + z) x3 + x2y + x2z + 2x2y + 2xy2 + 2xyz + 2x2z + 2xzy + 2xz2 + y2x + y3 + y2z +2yzx + 2y2z + 2yz2 + z2x + z2y + z3 Simplify x3 + 3x2y + 3x2z + 3xy2 + 6xyz + 3xz2 + y3 + 3y2z + 3yz2 + z3

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