Chapter 2 Combinatorial Analysis

1. Chapter 2Combinatorial Analysis 主講人:虞台文

2. Content • Basic Procedure for Probability Calculation • Counting • Ordered Samples with Replacement • Ordered Samples without Replacement Permutations • Unordered Samples without Replacement  Combinations • Binomial Coefficients • Some Useful Mathematic Expansions • Unordered Samples with Replacement • Derangement • Calculus

3. Chapter 2Combinatorial Analysis Basic Procedure for Probability Calculation

4. Basic Procedure for Probability Calculation • Identify the sample space . • Assign probabilities to certain events in A, e.g., sample point event P(). • Identify the events of interest. • Compute the desired probabilities.

5. Chapter 2Combinatorial Analysis Counting

6. Goal of Counting Counting the number of elements in a particular set, e.g., a sample space, an event, etc.

7. Cases • Ordered Samples w/ Replacement • Ordered Samples w/o Replacement • Permutations • Unordered Samples w/o Replacement • Combinations • Unordered Samples w/ Replacement

8. Chapter 2Combinatorial Analysis Ordered Samples with Replacement

9. Ordered Samples eat elements in samples appearing in different orders are considered different. ate tea

10. Ordered Samples w/ Replacement meet 1. Elements in samples appearing in different orders are considered different. 2. In each sample, elements are allowed repeatedly selected. teem mete

11.  k Ordered Samples w/ Replacement n distinct objects • Drawing k objects, their order is noted, among n distinct objects withreplacement. • The number of possible outcomes is

12. Example 1 How many possible 16-bit binary words we may have?  16 2 distinct objects

13. Example 2 Randomly Choosing k digits from decimal number, Find the probability that thenumber is a valid octal number. For any , P()=1/10k.

14. Chapter 2Combinatorial Analysis Ordered Samples without Replacement  Permutations

15. 清 以 心 也 可 Permutations 可以清心也 以清心也可 清心也可以 心也可以清 也可以清心

16.  k Ordered Samples w/o Replacement  Permutations n distinct objects • Drawing k objects, their order is noted, among n distinct objects withoutreplacement. • The number of possible outcomes is

17. Example 3 Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events? 1. Begin with an s. 2. Contains novowel. 3. Begins and ends with a consonant. 4. Contains onlyvowels.

18. E1: word begins with an s. E2: word contains no vowel. E3: word begins and ends with a consonant. E4: word contains only vowels. Define Example 3 P(E1)=? P(E2)=? P(E3)=? P(E4)=? Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events? 1. Begin with an s. 2. Contains novowel. 3. Begins and ends with a consonant. 4. Contains onlyvowels.

19. E1: word begins with an s. E2: word contains no vowel. E3: word begins and ends with a consonant. E4: word contains only vowels. Define Example 3 P(E1)=? P(E2)=? P(E3)=? P(E4)=? For any , P()=1/||.

20. Chapter 2Combinatorial Analysis Unordered Samples without Replacement  Combinations

21. Combinations ndistinct objects Choose k objects How many choices?

22. This notation is preferred Combinations • Drawing kobjects, their order is unnoted, among ndistinct objects w/o replacement, the number of possible outcomes is

23. More on

24. Examples

25. Example 4 • x • Denoting the all-assistant event as E, The mathematics department consists of 25 full professors, and 15 associate professors, and 35 assistant professors. A committee of 6 is selected at random from the faculty of the department. Find the probability that all the members of the committee are assistant professors.

26. Example 5 A poker hand hasfive cards drawn from an ordinary deck of 52 cards. Find theprobability that the poker hand has exactly2 kings. • x • Denoting the 2-king event as E,

27. m n 1 2 3 r 1 2 3 r Two boxes both have r balls numbered 1, 2, … , r. Two random samples of size m and n are drawn without replacement from the 1st and 2nd boxes, respectively. Find the probability that these two samples have exactly k balls with common numbers. Example 6 E P(“k matches”) = ? ||=? |E|=?

28. m n 1 2 3 r # possible outcomes from the 1st box. # possible k-matches. 1 2 3 r # possible outcomes from the 2nd box for each k-match. Example 6

29. m n 1 2 3 r 1 2 3 r Example 6

30. m n 1 2 3 r 1 2 3 r Example 6 樂透和本例有何關係?

31. m n 1 2 3 r 1 2 3 r Example 6 本式觀念上係由第一口箱子出發所推得

32. m n 1 2 3 r 1 2 3 r Example 6 觀念上，若改由第二口箱子出發結果將如何?

33. m n 1 2 3 r 1 2 3 r Example 6

34. m n 1 2 3 r 1 2 3 r Exercise

35. Chapter 2Combinatorial Analysis Binomial Coefficients

36. Binomial Coefficients

37. Binomial Coefficients

38. n terms Binomial Coefficients

39. n boxes Binomial Coefficients

40. Facts: Binomial Coefficients

41. Properties of Binomial Coefficients

42. Properties of Binomial Coefficients

43. Properties of Binomial Coefficients

44. Properties of Binomial Coefficients Exercise

45. Properties of Binomial Coefficients n不同物件任取k個  第一類取法:  第二類取法:

46. Properties of Binomial Coefficients Pascal Triangular

47. Properties of Binomial Coefficients Pascal Triangular 1 1 1 2 1 1 3 1 3 1 6 4 1 4 1

48. Properties of Binomial Coefficients 　吸星大法

49. Example 7-1

50. Fact: Example 7-2 kx+1 ?