Counting Product Rule

Transcription

1 Counting Product Rule Suppose a procedure can be broken down into a sequence of two tasks. If there are n 1 ways to do the first task and n 2 ways to do the second task, then there are n 1 * n 2 ways to do the procedure A x B = A B If A and B are finite sets, the number of elements in the Cartesian product of the sets is product of the # elements in each set Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

2 The Product Rule Example: How many different license plates can be made if each plate contains a sequence of three uppercase English letters followed by three digits? Solution: By the product rule, there are = 17,576,000 different possible license plates. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

3 Counting Suppose you have 4 shirts, 3 pairs of pants, and 2 pairs of shoes. How many different outfits do you have? Product Rule Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

4 Product Rule How many functions are there from set A to set B? A B 64 To define each function we have to make 3 choices, one for each element of A How many ways can each choice be made? Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

5 Product Rule How many one-to-one functions are there from set A to set B? A B 24 To define each function we have to make 3 choices, one for each element of A How many ways can each choice be made? Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

6 Counting Sum Rule If a task can be done either in one of n 1 ways or in one of n 2 ways, where none of the n 1 ways is the same as any of the set of n 2 ways, then there are n 1 + n 2 ways to do the task If A and B are disjoint sets then A U B = A + B Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

7 Counting Suppose a student can choose a computer project from one of three lists List A 23 projects List B 15 projects List C 19 projects No project is on more than one list How many possible projects are there to choose from? Sum Rule Just count them up = 57 Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

8 Decision Tree (Tree Diagrams) Suppose you have 4 shirts, 3 pairs of pants, and 2 pairs of shoes. How many different outfits do you have? Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

9 Decision Tree How many different best of 5 game series were possible between the Cubs and the Dogers? 20 A DT is a good model for a sequence of events. It assists in counting, and can help you see special structure in the problem. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

10 DNA and Genomes A gene is a segment of a DNA molecule that encodes a particular protein and the entirety of genetic information of an organism is called its genome. DNA molecules consist of two strands of blocks known as nucleotides. Each nucleotide is composed of bases: adenine (A), cytosine (C), guanine (G), or thymine (T). The DNA of bacteria has between 10 5 and 10 7 links (one of the four bases). Mammals have between 10 8 and links. So, by the product rule there are at least different sequences of bases in the DNA of bacteria and different sequences of bases in the DNA of mammals. The human genome includes approximately 23,000 genes, each with 1,000 or more links. Biologists, mathematicians, and computer scientists all work on determining the DNA sequence (genome) of different organisms. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

11 Telephone Numbering Plan Example: The North American numbering plan (NANP) specifies that a telephone number consists of 10 digits, consisting of a three-digit area code, a three-digit office code, and a four-digit station code. There are some restrictions on the digits. Let X denote a digit from 0 through 9. Let N denote a digit from 2 through 9. Let Y denote a digit that is 0 or 1. In the old plan (in use in the 1960s) the format was NYX-NNX-XXX. In the new plan, the format is NXX-NXX-XXXX. How many different telephone numbers are possible under the old plan and the new plan? Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

12 Telephone Numbering Plan Example: The North American numbering plan (NANP) specifies that a telephone number consists of 10 digits, consisting of a three-digit area code, a three-digit office code, and a four-digit station code. There are some restrictions on the digits. Let X denote a digit from 0 through 9. Let N denote a digit from 2 through 9. Let Y denote a digit that is 0 or 1. In the old plan (in use in the 1960s) the format was NYX-NNX-XXXX. In the new plan, the format is NXX-NXX-XXXX. How many different telephone numbers are possible under the old plan and the new plan? Solution: Use the Product Rule. There are = 160 area codes with the format NYX. There are = 800 area codes with the format NXX. There are = 640 office codes with the format NNX. There are = 10,000 station codes with the format XXXX. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

13 Telephone Numbering Plan Example: The North American numbering plan (NANP) specifies that a telephone number consists of 10 digits, consisting of a three-digit area code, a three-digit office code, and a four-digit station code. There are some restrictions on the digits. Let X denote a digit from 0 through 9. Let N denote a digit from 2 through 9. Let Y denote a digit that is 0 or 1. In the old plan (in use in the 1960s) the format was NYX-NNX-XXXX. In the new plan, the format is NXX-NXX-XXXX. How many different telephone numbers are possible under the old plan and the new plan? Solution: Use the Product Rule. There are = 160 area codes with the format NYX. There are = 800 area codes with the format NXX. There are = 640 office codes with the format NNX. There are = 10,000 station codes with the format XXXX. Number of old plan telephone numbers: ,000 = 1,024,000,000. Number of new plan telephone numbers: ,000 = 6,400,000,000. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

14 Inclusion Exclusion Principle If a task can be done either in one of n 1 ways or in one of n 2 ways, but some of the n 1 ways are the same as some of the n 2 ways. Are there n 1 + n 2 ways to do the task now? If sets A and B are NOT disjoint A U B = A + B - A n B Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

15 Example Count the number of bit strings of length 4 which begin with a 1 or end with 00 How many bit strings of length 4 begin with a 1? 2 3 = 8 How many bit strings end in 00? 2 2 = 4 So the total is = 12? How many bit strings end in 00 and start with 1? 2 The real total is = 10 Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

16 Combining the Sum and Product Rule Example: Suppose statement labels in a programming language can be either a single letter or a letter followed by a digit. Find the number of possible labels. Solution: Use the product rule = 286 Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

17 Counting Passwords Combining the sum and product rule allows us to solve more complex problems. Example: Each user on a computer system has a password, which is six to eight characters long, where each character is an uppercase letter or a digit. Each password must contain at least one digit. How many possible passwords are there? Solution: Let P be the total number of passwords, and let P 6, P 7, and P 8 be the passwords of length 6, 7, and 8. By the sum rule P = P 6 + P 7 +P 8. To find each of P 6, P 7, and P 8, we find the number of passwords of the specified length composed of letters and digits and subtract the number composed only of letters. We find that: P 6 = =2,176,782, ,915,776 =1,867,866,560. P 7 = = 78,364,164,096 8,031,810,176 = 70,332,353,920. P 8 = = 2,821,109,907, ,827,064,576 =2,612,282,842,880. Consequently, P = P 6 + P 7 +P 8 = 2,684,483,063,360. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

18 Internet Addresses Version 4 of the Internet Protocol (IPv4) uses 32 bits. Class A Addresses: used for the largest networks, a 0,followed by a 7- bit netid and a 24-bit hostid. Class B Addresses: used for the medium-sized networks, a 10,followed by a 14-bit netid and a 16-bit hostid. Class C Addresses: used for the smallest networks, a 110,followed by a 21-bit netid and a 8-bit hostid. Neither Class D nor Class E addresses are assigned as the address of a computer on the internet. Only Classes A, B, and C are available is not available as the netid of a Class A network. Hostids consisting of all 0s and all 1s are not available in any network. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

19 Counting Internet Addresses Example: How many different IPv4 addresses are available for computers on the internet? Solution: Use both the sum and the product rule. Let x be the number of available addresses, and let x A, x B, and x C denote the number of addresses for the respective classes. To find, x A : = 127 netids = 16,777,214 hostids. x A = ,777,214 = 2,130,706,178. To find, x B : 2 14 = 16,384 netids = 16,534 hostids. x B = 16,384 16, 534 = 1,073,709,056. To find, x C : 2 21 = 2,097,152 netids = 254 hostids. x C = 2,097, = 532,676,608. Hence, the total number of available IPv4 addresses is x = x A + x B + x C = 2,130,706, ,073,709, ,676,608 = 3, 737,091,842. Not Enough Today!! The newer IPv6 protocol solves the problem of too few addresses. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

20 Pigeonhole Principle If n pigeons fly into k pigeonholes and k < n, then some pigeonhole contains at least two pigeons. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

21 Pigeonhole Principle We can use this simple little fact to prove amazingly complex things. If n pigeons fly into k pigeonholes and k < n, then some pigeonhole contains at least two pigeons. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

22 The Pigeonhole Principle Pigeonhole Principle: If k is a positive integer and k + 1 objects are placed into k boxes, then at least one box contains two or more objects. Proof: We use a proof by contraposition. Suppose none of the k boxes has more than one object. Then the total number of objects would be at most k. This contradicts the statement that we have k + 1 objects. Corollary 1: A function f from a set with k + 1 elements to a set with k elements is not one-to-one. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

23 Pigeonhole Principle Example: Among any group of 367 people, there must be at least two with the same birthday, because there are only 366 possible birthdays. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

24 The Generalized Pigeonhole Principle 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. Proof: We use a proof by contraposition. Suppose that none of the boxes contains more than N/k 1 objects. Then the total number of objects is at most where the inequality N/k < N/k + 1 has been used. This is a contradiction because there are a total of n objects. Example: Among 100 people there are at least 100/12 = 9 who were born in the same month. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

25 Pigeonhole Principle Let S contain any 6 positive integers. Then, there is a pair of numbers in S whose difference is divisible by 5. Let S = {a 1,a 2,a 3,a 4,a 5,a 6 }. Each of these has a remainder when divided by 5. What can these remainders be? 6 numbers, 5 possible remainders what do we know? 0, 1, 2, 3, or 4 Some pair has the same remainder, by PHP Consider that pair, a i and a j, and their remainder r. a i = 5m + r, and a j = 5n + r. Their difference: a i - a j = (5m + r) - (5n + r) = 5m - 5n = 5(m-n), which is divisible by 5. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

26 Pigeonhole Principle Six people go to a party. Either there is a group of 3 who all know each other, or there is a group of 3 who are all strangers. Consider one person. She either knows or doesn t know each other person. But there are 5 other people! So, she knows, or doesn t know, at least 3 others. Let s say she knows 3 others. If any of those 3 know each other, we have a blue ê, which means 3 people know each other. If not, then those 3 people must be strangers. But then we ve proven our conjecture for this case. The case where she doesn t know 3 others is similar. Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

27 Extra Examples In questions below suppose that a word is any string of seven letters of the alphabet, with repeated letters allowed. 1) How many words are there? 2) How many words end with the letter T? 3) How many words begin with R and end with T? 4) How many words begin with A or B? 5) How many words begin with A or end with B? 6) How many words begin with A or B and end with A or B? Extensible Networking Platform CSE 240 Logic and Discrete Mathematics

28 Extra Examples Solutions In questions below suppose that a word is any string of seven letters of the alphabet, with repeated letters allowed. 1) How many words are there? ) How many words end with the letter T? ) How many words begin with R and end with T? ) How many words begin with A or B? 2 * ) How many words begin with A or end with B? ) How many words begin with A or B and end with A or B? 4 * 26 5 Extensible Networking Platform CSE 240 Logic and Discrete Mathematics