King ABDUL AZIZ University Faculty Of Computing and Information Technology CPCS 222 Discrete Structures I Counting Dr. Eng. Farag Elnagahy farahelnagahy@hotmail.com Office Phone: 67967
The Basics of counting Combinatorics is the mathematics of counting and arranging objects. Counting of objects with certain properties (enumeration) is required to solve many different types of problems. For example, counting is used to: Determine number of ordered or unordered arrangement of objects. Generate all the arrangements of a specified kind which is important in computer simulations. Compute probabilities of events. Analyze the chance of winning games, lotteries etc. Determine the complexity of algorithms. 2
The Basics of counting Two basic counting principles The Sum Rule The Product Rule Let us consider two tasks: m is the number of ways to do task1 n is the number of ways to do task2 Performing task1 does not accomplish task2 and vice versa (task1 and task2 are independent of each other). Sum rule: the number of ways that either task1 or task2 can be done, but not both, is m+n. Product rule: the number of ways that both task1 and task2 can be done in mn. 3
The Basics of counting (Examples) Example: A student can choose a computer project from one of three lists. The three lists contain 23, 15, and 19 possible projects respectively. How many possible projects are there to choose from? (23+15+19=57) Example: The chairs of an auditorium are to be labeled with a letter and a positive integer not to exceed 100. What is the largest number of chairs that can be labeled differently? (26x100=2600) 4
The Basics of counting (Set version) If A is the set of ways to do task1, and B the set of ways to do task2, and if A and B are disjoint, then: The ways to do either task1 or task2 are A B, and A B = A + B The ways to do both task1 and task2 are A B, and A B = A B The number of different subsets of a finite set(s) is? 000000000000 100000000000 010000000000 2 s. 111111111111 5
The Basics of counting (Examples) What is the value of k after the following code has been executed? K:=0 for i 1 :=1 to n 1 K:= K +1 for i 2 =1 to n 2 K:= K +1 for i 3 =1 to n 3 K:= K +1 K:=0 for i 1 :=1 to n 1 for i 2 :=1 to n 2 for i 3 :=1 to n 3 K:= K +1 K= n 1 + n 2 + n 3 K= n 1 x n 2 x n 3 6
The Basics of counting (Examples) Count the number of print statements in this algorithm: for i := 1 to n begin for j := 1 to n print hello for k := 1 to n print hello end The total number of print statements executed is n (n+n) = 2n 2. 7
The Basics of counting (Examples) Count the number of print statements in this algorithm: for i := 1 to n begin for j := 1 to i print hello for k := i + 1 to n print hello end for each i, the number of print statements executed is i in the j loop plus n i in the k loop. Therefore, for each i, the number of print statements is i + (n i) = n. Therefore the total number of print statements executed is n n = n 2. 8
The Basics of counting (Examples) In a computer language The name of a variable is a string of one or two alphanumeric characters. uppercase and lowercase letters are not distinguished. 26 English letter, 10 digits. the variable name must begin with letter. there are five strings of two characters that are reserved for programming use. How many different variable names are there? V 1 string (one character),v 2 string (two characters) V=V 1 +V 2 V 1 =26 V 2 =26x(26+10) 5 =26x36 5 =931 V=26+931=957 9
The Basics of counting (Examples) How many different license plates are available if each plate contains a sequence of three letters followed by three digits? L 1 L 2 L 3 D 1 D 2 D 3 Each of the three letters can be written in 26 different ways, and each of the three digits can be written in 10 different ways. Hence, by the product rule, there is a total of 26 26 26 10 10 10 = 17,576,000 different license plates possible. 10
The Basics of counting (Examples) Each user on a computer system has a password, which is six to eight characters long, where each character is an uppercase letter or digit. Each password must contain at least one digit. How many passwords are there? (string includes Letters & Digits)- (string with no digits) P=P 6 +P 7 +P 8 P 6 =36 6-26 6 =1 867 866 560 P 7 =36 7-26 7 =70 332 353 920 P 8 =36 8-26 8 =2 612 282 842 880 P=P 6 +P 7 +P 8 =2 684 483 063 360 11
The Basics of counting (Examples) In version 4 of the Internet Protocol (IPv4) The internet address is a string of 32 bits as follows: Class A address [ 0 -netid(7 bits) hostid (24 bits)] Class B address [ 10 -netid(14 bits) hostid (16 bits)] Class C address [ 110 -netid(21 bits) hostid (8 bits)] Class D address [ 1110 multicast address (28 bits)] Class E address [ 11110 - address (27 bits)] Where, Network number(netid) -host number(hostid) Restrictions 1111111 is unavailable in netid All 0s and all 1s are unavailable in hostid The computer on the Internet has either class A or B or C addresses. How many different IPv4 addresses are available for computers on the Internet? (2 7-1)(2 24-2)+(2 14-1)(2 16-2)+(2 21-1)(2 8-2) 12
The Basics of counting (Examples) How many bit string of length eight either start with a 1 bit or end with the two bits 00? 1 - - - - - - - 2 7 =128 ways - - - - - - 0 0 2 6 =64 ways 1 - - - - - 0 0 2 5 =32 ways 128 + 64-32= 160 13
The Basics of counting (Examples) A computer company receives 350 applications from computer graduates for a job. Suppose that 220 of these people majored in CS, 147 majored in business, and 51 majored both in CS and in business. How many of these applicants majored neither in CS nor in business? Let A 1 be the set of students who majored in CS Let A 2 be the set of students who majored in business The number of students who majored either in CS or in business (or both) is A 1 A 2 = A 1 + A 2 - A 1 A 2 =220+147-51=316 The number of applicants who majored neither in CS nor in business is 350-316=34 14
The Basics of counting (Examples) Exercises PP. 344-347 1-12 15
The Pigeonhole principle The Pigeonhole principle Suppose there are n pigeons, k pigeonholes, and n>k. If these n pigeons fly into these k pigeonholes, then some pigeonhole must contain at least two pigeons. If k+1 objects are assigned to k places, then at least 1 place must be assigned 2 objects. 7 pigeons 6 pigeonholes 16
The Pigeonhole principle In terms of the assignment function: If f:a B and A B +1, then some element of B has 2 pre-images under f.( f is not one-to-one) How many students must be in class to guarantee that at least two students receive the same score on the final exam, if the exam is graded on a scale from 0 to 100 points? Greater than 101 17
The Pigeonhole principle (Examples) The generalized Pigeonhole principle If N objects are placed into k boxes, then there is at least one box containing at least N/K objects. e. g., there are N=280 students in this class. There are k=52 weeks in the year. Therefore, there must be at least 1 week during which at least 280/52 = 5.38 =6 students in the class have a birthday. 18
The Pigeonhole principle (Examples) There are 280 students in the class. Without knowing anybody s birthday, what is the largest value of n for which we can prove that at least n students must have been born in the same month? 280/12 = 23.3 = 24 What is the minimum number of students required in a discrete math class to be sure that at least six will receive the same grade, if there are five possible grades, A, B, C, D, and F? 19
Permutations and Combinations Permutations A. In how many ways can we select 3 students from a group of 5 students to stand in line for a picture? B. In how many ways can we arrange all 5 of these students in a line for a picture? Note that the order in which we select the students matters. 5 ways to select the first student 4 ways to select the second student 3 ways to select the third student 2 ways to select the fourth student 1 way to select the fifth student A. 5.4.3=60 B. 5.4.3.2.1=120 20
Permutations and Combinations Permutations A permutation of a set S of objects is an ordered arrangement of the elements of S where each element appears only once: e.g., 1 2 3, 2 1 3, 3 1 2 An ordered arrangement of r distinct elements of S is called an r-permutation. The number of r-permutations of a set S with n= S elements is P(n,r) = n(n 1)(n 2) (n r+1) = n!/(n r)! 21
Permutations and Combinations Permutations (examples) How many ways are there to select a first-prize winner, a second-prize winner,and a third-prize winner from 100 different people who have entered a contest? P(100,3)=100.99.98=970200 How many permutations of the letters ABCDEFGH contain the string ABC? ABC, D, E, F, G, H we have 6 objects Theses object can occur in any order 6.5.4.3.2.1 There are 6!=720 permutations 22
Permutations and Combinations Combinations The number of ways of choosing r elements from S (order does not matter). S={1,2,3} e.g., 1 2, 1 3, 2 The number of r-combinations C(n,r) of a set with n= S elements is 23
Permutations and Combinations Example S={1,2,3}, all permutations={(1,2,3),(2,1,3),(1,3,2),(2,3,1),(3,1,2),(3,2,1 )} all 2-permutations={(1,2),(2,1),(1,3),(3,1),(2,3),(3,2)} P(3,3)=3*2*1=6, P(3,2)=3*2=6 S={1,2,3}, all 2-combinations={{1,2},{1,3},{2,3}} Comparing to all 2-permutations, we see we ignore order, 24
Permutations and Combinations Example How many ways are there to choose a committee of size five consisting of three women and two men from a group of ten women and seven men? The number of ways to choose three women is C(10, 3) The number of ways to choose two men is C(7, 2). Using the product rule to choose three women and two men, the answer is C(10, 3) C(7, 2) = 2, 520. 25
Permutations and Combinations Example A class has 20 women and 16 men. In how many ways can you (a) Put all the students in a row? (b) Put 7 of the students in a row? (c) Put all the students in a row if all the women are on the left and all the men are on the right? Solution: (a) There are 36 students. They can be put in a row in 36! ways. (b) You need to have an ordered arrangement of 7 out of 36 students. The number of such arrangements is P(36, 7). (c) You need to have an ordered arrangement of all 20 women AND and ordered arrangement of all 16 men. By the product rule, this can be done in 20! 16! ways. 26
Permutations and Combinations Exercises PP. 360-362 1 4 5 7 27