1. Unit VI Discrete Structures Permutations and Combinations SE (Comp.Engg.)

2. BOTH PERMUTATIONS AND COMBINATIONS USE A COUNTING METHOD CALLED FACTORIAL

3. Lets start with a simple example. A student is to roll a die and flip a coin. How many possible outcomes will there be?

4. 1H 2H 3H 4H 5H 6H 1T 2T 3T 4T 5T 6T

5. The number of ways to arrange the letters ABC: ____ ____ ____ Number of choices for first blank ? 3 ____ ____ 3 2 ___ Number of choices for second blank? Number of choices for third blank? 3 2 1

6. Permutations A Permutation is an arrangement of items in a particular order. The number of Permutations of n items chosen r at a time, is given by the formula

7. Arrange 2 alphabets from a,b,c. ab, ba, ac,ca,bc,cb 3P 2 =3!/(3-2)!=6

8. A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?

10. CIRCULAR PERMUTATIONS When items are in a circular format, to find the number of different arrangements, divide: n! / n

11. Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there?

12. 6!  6 = 120

13. Combination A Combination is an arrangement of items in which order does not matter. The number of Combinations of n items chosen r at a time, is given by the formula

14. To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible?

16. A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions?

18. Product and Sum Rules Product Rule: If we need to perform procedure 1 AND procedure 2. There are n 1 ways to perform procedure 1 and n 2 ways to perform procedure 2. There are n 1n 2 ways to perform procedure 1 AND procedure 2.

19. Sum Rule: If need to perform either procedure 1 OR procedure 2. There are n 1 ways to perform procedure 1 and n 2 ways to perform procedure 2. There are n 1 +n 2 ways to perform procedure 1 OR procedure 2. This “OR” is an “exclusive OR.” One choice or the other, but not both.

20. How many vehicle number plates can be made if each plate contains two different letters followed by three different digits

21. Two different letters are made in 26P 2 ways. 3 different digits are combined in 10P 3 ways Total no. of number plates= 26P 2 * 10P 3 = 26!*10!/(24!*7!) =26*25*10*9*8 =468000

22. An 8 member team is to be formed from a group of 10 men and 15 women. In how many ways can the team be chosen if : (i) The team must contain 4 men and 4 women (ii) There must be more men than women (iii) There must be at least two men

23. The team must contain 4 men and 4 women = 286650 (ii) There must be more men than women = (iii) There must be at least two men

24. Passwords consist of character strings of 6 to 8 characters. Each character is an upper case letter or a digit. Each password must contain at least one digit. How many passwords are possible?

25. Total number is # passwords with 6 char. + # passwords with 7 char. + # pws 8 char. (=P6+P7+P8).

26. P6: # possibilities without constraint : 36 6 # exclusions is # passwords without any digits is 26 6 And so, P6 = 36 6 -26 6

27. Similarly, P7 = 36 7 -26 7 and P8 = 36 8 -26 8

28. P = P6+P7+P8 = 36 6 -26 6 + 36 7 -26 7 + 36 8 -2

29. How many bit-strings of length 8 either begin with 1 or end with 00?

30. A = 8-bit strings starting with 1 |A| = # of 8-bit strings starting with 1 is 2 7 B = 8-bit strings starting with 00 |B| = # of 8-bit strings ending with 00 is 2 6 # of bit-strings begin with 1 and end with 00 is 2 5.

31. # of 8-bit strings starting with 1 or ending with 00 is 2 7 + 2 6 - 2 5

32. |A  B| = |A| + |B| - |A  B| Inclusion–Exclusion Principle

33. 33 Permutations with non-distinguishable objects The number of different permutations of n objects, where there are non-distinguishable objects of type 1, non-distinguishable objects of type 2, …, and non-distinguishable objects of type k, is i.e., C(n, )C(n-, )…C(n- - -…-, )

34. How many different strings can be made by reordering the letters of the word OFF

35. 3!/2!=3 OFF FFO FOF

36. ONE OEN NEO NOE ENO EON

37. How many different strings can be made by reordering the letters of the word SUCCESS

38. Generating Permutations Lexicographic method

39. For the following 4 combinations from the set f= {1;2;3;4;5;6;7} find the combination that immediately follows them in lexicographic order 1234 is followed by 3467 is followed by 4567 is followed by

40. What is probability? Probability is the measure of how likely an event or outcome is. Different events have different probabilities!

41. How do we describe probability? You can describe the probability of an event with the following terms: – certain (the event is definitely going to happen) – likely (the event will probably happen, but not definitely) – unlikely (the event will probably not happen, but it might) – impossible (the event is definitely not going to happen)

42. probabilities are expressed as fractions. – The numerator is the number of ways the event can occur. – The denominator is the number of possible events that could occur.

43. 6.43 Random Experiment… …a random experiment is an action or process that leads to one of several possible outcomes. For example: ExperimentOutcomes Flip a coinHeads, Tails Exam MarksNumbers: 0, 1, 2,..., 100 Assembly Timet > 0 seconds Course GradesF, D, C, B, A, A+

44. What is the probability that I will choose a red marble? In this bag of marbles, there are: – 3 red marbles – 2 white marbles – 1 purple marble – 4 green marbles

45. Probability example Sample space: the set of all possible outcomes. Probabilities: the likelihood of each of the possible outcomes (always 0  P  1.0).

46. 6.46 Probabilities… List the outcomes of a random experiment… This list must be exhaustive, i.e. ALL possible outcomes included. Die roll {1,2,3,4,5}Die roll {1,2,3,4,5,6} The list must be mutually exclusive, i.e. no two outcomes can occur at the same time: Die roll {odd number or even number} Die roll{ number less than 4 or even number}

47. A and B are independent if and only if P(A&B)=P(A)*P(B) A and B are mutually exclusive events: P(A or B) = P(A) + P(B)

48. 6.48 Events & Probabilities… The probability of an event is the sum of the probabilities of the simple events that constitute the event. E.g. (assuming a fair die) S = {1, 2, 3, 4, 5, 6} and P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6 Then: P(EVEN) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2

49. Female Low Male e.g. 1. The following table gives data on the type of car, grouped by petrol consumption, owned by 100 people. 42123 73312 HighMedium One person is selected at random. L is the event “the person owns a low rated car” F is the event “a female is chosen”. Find (i) P (L) (ii) P (F  L)(iii) P (F| L) 100 Total

50. (i) P (L) = Solution: Find (i) P (L) (ii) P (F  L)(iii) P (F L) 100 42123Female 73312Male HighMediumLow (ii) P (F  L) = Total (iii) P (F L)  Notice that P (L)  P (F L) So, P (F  L) = P(F|L)  P (L) = P (F  L)

51. 45 R F e.g. 2. I have 2 packets of seeds. One contains 20 seeds and although they look the same, 8 will give red flowers and 12 blue. The 2nd packet has 25 seeds of which 15 will be red and 10 blue. Draw a Venn diagram and use it to illustrate the conditional probability formula. Solution: 15 10 12 P (R  F) = P (F) =P (R F) = 8 P (R  F) = P(R|F)  P (F) So, P (R F)  P (F) = LetR be the event “ Red flower ” and F be the event “ First packet ”

52. Bayes’ Rule: derivation Definition: Let A and B be two events with P(B)  0. The conditional probability of A given B is:

54. Example Three jars contain colored balls as described in the table below. – One jar is chosen at random and a ball is selected. If the ball is red, what is the probability that it came from the 2 nd jar? Jar #RedWhiteBlue 1341 2123 3432

55. Example We will define the following events: – J 1 is the event that first jar is chosen – J 2 is the event that second jar is chosen – J 3 is the event that third jar is chosen – R is the event that a red ball is selected

56. Example The events J 1, J 2, and J 3 mutually exclusive – Why? You can’t chose two different jars at the same time Because of this, our sample space has been divided or partitioned along these three events

57. Venn Diagram Let’s look at the Venn Diagram

58. Venn Diagram All of the red balls are in the first, second, and third jar so their set overlaps all three sets of our partition

59. Finding Probabilities What are the probabilities for each of the events in our sample space? How do we find them?

60. Computing Probabilities Similar calculations show:

61. Venn Diagram Updating our Venn Diagram with these probabilities:

62. Where are we going with this? Our original problem was: – One jar is chosen at random and a ball is selected. If the ball is red, what is the probability that it came from the 2 nd jar? In terms of the events we’ve defined we want:

63. Finding our Probability