1. 1 Block 3 Discrete Systems Lesson 9 – Methods of Discrete Counting You can count on this presentation. Narrator: Charles Ebeling

2. 2 Block 3 – Discrete Systems Lesson 9 – Methods of Discrete Counting Lesson 10 - Sequences and Series Lesson 11 – Discrete Probability Models Lesson 12 – Discrete Optimization Models

3. 3 What are Discrete Systems? Discrete mathematics, sometimes called finite mathematics, is the study of mathematical structures that are fundamentally discrete, in the sense of not supporting or requiring the notion of continuity. Most, if not all, of the objects studied in finite mathematics are countable sets, such as the integers. Discrete systems is the application of discrete mathematics to solving certain types of problems by constructing discrete models and applying specific algorithms to solve them.

4. 4 Topics in Discrete Mathematics Logic - a study of reasoning Set theory - a study of collections of elements Number theory Combinatorics - a study of counting Graph theory Algorithmic - a study of methods of calculation Information theory The theory of computability and complexity - a study on theoretical limitations on algorithms Probability theory Markov chains

5. 5 Combinatorics Combinatorics is a branch of mathematics that studies collections (usually finite) of objects that satisfy specified criteria. In particular, it is concerned with "counting" the objects in those collections (enumerative combinatorics). I can count. Listen! 1, 2, 3, …

6. 6 Fundamental Principle of Counting If something can be chosen, or can happen, or be done, in m different ways, and, after that has happened, something else can be chosen in n different ways, then the number of ways of choosing both of them is m · n. a, b, c, d e, f, g, h 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 select a letterselect a number 8 x 12 = 96 outcomes

7. 7 Another example For example, the letters a, b, c, d are put into a hat and two of them are drawn in succession. We can draw the first in 4 different ways: either a or b or c or d. After that has happened, there are 3 ways to choose the second. Therefore, there are 4· 3 or 12 possible ways to choose two letters from four in order.

8. 8 Permutations A PERMUTATIONS of n objects, is all of their possible arrangements or orderings. For example, the permutation of the letters a, b, and c is: abc acb bac bca cab cba Of primary interest is knowing the number of permutations that are available from a collection of objects. As the number of things (n) increases, their permutations grow astronomically. For example, if twelve different things are permuted, then the number of their permutations is 479,001,600.

9. 9 Permutations To find the number of permutations of four objects: there are 4 ways to choose the first, 3 ways remain to choose the second, 2 ways to choose the third, and 1 way to choose the last. Therefore the number of permutations of 4 different things is 4· 3· 2· 1 = 24 The number of permutations of n different things taken all at a time is n!.

10. 10 Permutations of less than all A permutation of n things taken r at a time is an ordered arrangement of r objects selected from the n objects. Given the letters a, b, c then the permutations of the 3 letters taken 2 at a time are: ab ba ac ca bc cb

11. 11 How many are there? The number of permutations of n objects taken r at a time is denoted by P(n,r) or n P r or P n,r. Example: Given the 4 letters a, b, c, and d, how many permutations are there if taken 3 at a time? answer: 4 choices for the 1 st letter, 3 for the 2 nd, and 2 for the 3 rd. Therefore P(4,3) = (4)(3)(2) = 4! / 1! = 24 2 at a time? ans. P(4,2) = (4)(3) = 4!/2! = 12

12. 12 Permutations with Repetitions The number of permutations of n objects of which n 1 are alike, n 2 are alike, …, n r are alike is Example: How many distinct signals, each consisting of 6 flags hung in a vertical line, can be formed from 4 identical red flags, and 2 blue flags?

13. 13 Ordered Sampling When (random) sampling r objects from a collection of n objects where the order in which the objects are selected is important then the number of such order samples is given by: P(n,r) if sampling without replacement n r if sampling with replacement Why sampling without replacement is just an r-permutation!

14. 14 Sampling without replacement How many different ways can an ordered dozen apples be selected from a bushel of 84 apples? This one is for my professor.

15. 15 Sampling with replacement In the state lottery 100 balls containing the numbers 00 to 99 are placed in a cylindrical container which is then rotated. A ball is selected at random, the number recorded, and then placed back into the cylinder. This is repeated 5 times in order to randomly form a 5-number winning sequence. How many different numbers are possible? n = 100, r = 5; therefore n r = 100 5 = 10,000,000,000

16. 16 Combinations A combination is an unordered selection of objects. A combination of n objects taken r at a time is any selection of r of the objects where order does not count. The number of combinations of n objects taken r at a time is denoted by

17. 17 The Number of Combinations Since an ordered arrangement of r objects from among n objects is a permutation where then to ignore the order of the r objects, divide by r!, the number of ways in which each set of r objects may be ordered.

18. 18 Proof by Example CombinationsPermutations abcabc, acb, bac, bca, cab, cba abdabd, adb, bad, bda, dab, dba acdacd, adc, cad, cda, dac, dca bcdbcd, bde, cbd, cdb, dbc, dcb Given 4 objects (a,b,c,d) taken 3 at a time

19. 19 Permutations versus Combinations The permutation of a number of objects is the number of different ways they can be ordered; i.e. which is first, second, third, etc. If you wish to choose some objects from a larger number of objects, the way you position the chosen objects is also important. With combinations, on the other hand, one does not consider the order in which objects were chosen or placed, just which objects were chosen. Permutations - position important (although choice may also be important) Combinations - chosen important,

20. 20 Some Combination Examples How many committees of 3 can be formed from 8 people? C(8,3) = 8! /(5! 3!) = (8)(7)(6) / [(1)(2)(3)] = 56 In a deck of 52 playing cards, how many different poker hands of 5 cards are there? C(52,5) = 2,598,960 In a deck of 52 playing cards, how many ways can a spade flush (5 spades) be dealt? C(13,5) = 1287 How many 4 letter words can be formed from the English alphabet if at least one vowel (not y) must be present? C(5,1) 26 3 = (5) (2600) = 87,880

21. 21 Ordered Partitions A partition of a set X is a subdivision of X into subsets which are disjoint and whose union is X. Example: An urn contains 7 balls numbered one through seven. How many ways can we draw first 2 balls from the urn, then 3 balls from the urn, and lastly 2 balls from the urn? Compute the number of ordered partitions (A 1, A 2, A 3 ) of the set of 7 balls into cells A 1 containing 2 balls, A 2 containing 3 balls, and A 3 containing 2 balls. ordered because { (1,2), (3,4,5), (6,7)} and { (6,7), (3,4,5), (1,2) } are different outcomes. The number of such ordered partitions =

22. 22 Need another example? Joe the plumber has 12 house calls to make over the next 3 days. How many different ways can 4 customers be scheduled each day? The order of service each day is unimportant. Knowing how many ways I can schedule my customers has really helped the business. The business was in the toilet

23. 23 The Grab Bag An assortment of counting problems that will delight your fancy

24. 24 What’s behind the door? A door can be opened only with a security code that consists of five buttons: 1, 2, 3, 4, 5. A code consists of pressing any one button, or a) any two, or any three, or any four, or all five. How many possible codes are there? (You are to press all the buttons at once, so the order doesn't matter.) This is the sum of all the combinations of 5 things – except not taking any, 5 C 0, which is 1. The sum of all those combinations, then, is 2 5 − 1 = 32 − 1 = 31.

25. 25 An Interesting Relationship Look, two ways to compute the same value.

26. 26 Still trying to open that door If, to open the door, you must press three codes (with 5 buttons available), then how many possible ways are there to open the door? Assume that the same code may be repeated. There are 31 ways to choose the first code. Again, 31 ways to choose the second, and 31 ways to choose the third. Therefore, the total number of ways to open the door is 31 3 = 29,791.

27. 27 Shift Scheduling The manager of a small plant wishes to determine the number of ways he can assign men to the night shift in order to establish a rotation schedule. He has 15 employees that can serve as operators of the production equipment, 8 that are fully qualified maintenance technicians, and 4 are shift supervisors. The night shift requires 6 operators, 2 maintenance technicians, and one supervisor. But I like the night shift!

28. 28 Merry Matchmaking – a Valentine’s day special MerryMatchmaking.com is an on-line matchmaking service that attempts to introduce couples to one another based upon a compatibility index that is computed for each potential couple. If there are currently 22 males and 17 females that have registered for this service, how many compatibility indices (pairwise comparisons) will have to be computed? 22 x 17 = 374

29. 29 More Merry Matchmaking If sex is ignored, how many compatibility indices (pairwise comparisons) will have to be computed? There are currently 22 males and 17 females that have registered for this service. C(39,2) = (39)(38)/2 =1482/2 = 741 Another Merry Matchmaking success.

30. 30 The Traveling Salesman A salesman living in Dayton, Ohio is planning his next road trip where he will visit 8 different cities and then return to Dayton. He wishes to find the route that will minimize the total distance he must travel. How many routes are possible assuming he can travel from any one city to any of the other cities directly? P(8,8)=8! = 40,320

31. 31 Problems in the Lab By accident a chemist combined three laboratory substances that yielded a desirable product. Unfortunately the lab assistant did not record the names of the chemicals that were combined. There are 40 substances available in the lab. If the three chemicals in question must be determined by successive trial and error experiments, what is the maximum number of tests that might be made? Did I record the ingredients? Oh well, there can’t be too many combinations to try.

32. 32 Back to the Lab A known catalyst was used in the accidental reaction. Because of this the order on which the three ingredients are mixed is important. What now is the maximum number of experiments that might be made? Oh no!

33. 33 The Assignment Problem There are n workers and n tasks to be performed. The time it takes worker i to perform task j is c ij. Which task should be assigned to which workers? I want the easy task.

34. 34 How many solutions? If there are n workers and n tasks there are n! (factorial) possible assignments. Example: Workers are Al, Art, Alice, and Ann. There are four tasks: 1,2,3, & 4. 4321 = 4! = 24 AlArtAliceAnn If n = 10 then 10! = 3,628,800 if n = 20 then 20! = 2,432,902,008,176,640,000

35. 35 How long to solve – 20 workers? Assume each solution can be evaluated in.01 sec; then.01 x 2,432,902,008,176,640,000 sec. = 24,329,020,081,766,400 / 3600 hrs = 6,758,061,133,824 / 24 days = 281,585,880,576 / 365 yrs = 771,468,166 years Not in my lifetime! seconds hours days years

36. 36 The Methods of Combinatorics has Concluded - Now go and Practice Your Counting!