1. 1 Section 4.1 Basics of Counting

2. 2 Basic Counting Principles: Sum Rule Suppose you have two tasks to perform: –The first task can be done in n 1 ways; –The second task can be done in n 2 ways; –If the two tasks can’t be done simultaneously, then there are n 1 + n 2 ways to do either task Application of sum rule is pretty obvious: add up the number of variations to arrive at the total, which is your answer Can be extended to more than 2 tasks

3. 3 Example 1 (using sum rule) Suppose one CS major or one engineering major is eligible for the prestigious Sheller scholarship Suppose there are 20 CS majors and 15 engineering majors to choose from - how many choices are there for the scholarship? 20 + 15 = 35 possible choices

4. 4 Example 2 (using sum rule) What is the value of k after the following code is executed? k = 0; for (int i 1 =0; i 1 <n 1 ; i 1 ++) k++; for (int i 2 =0; i 2 <n 2 ; i 2 ++) k++;. for (int i m =0; i m <n m ; i m ++) k++;

5. 5 Example 2 Initial value of k is 0 There are m loops Each iteration adds 1 to k Let T i = task of traversing the ith loop; can be done n i ways, since traversal is done n i times Since no 2 tasks can be done at once, final value of k, which is the number of ways to do T i, where i = 1 … m is: n 1 + n 2 + … + n m

6. 6 Sum rule & sets If A 1, A 2, …, A m are disjoint sets: A 1  A 2  …  A m =  Then the number of elements in the union of these sets = the sum of the number of elements in each set: |A 1  A 2  …  A m | = |A 1 | + |A 2 | + … + |A m |

7. 7 Basic Counting Principles: Product Rule Suppose a procedure can be broken down into 2 tasks, with n 1 ways to do the first task and n 2 ways to do the second task After the first task is done, there are n 1 n 2 ways to the do the procedure Can extend product rule as sum rule was extended - in a procedure consisting of tasks T 1, T 2, … T m, each of which can be done n ways, there are n 1 *n 2 * … *n m ways to carry out the procedure

8. 8 Example 3 (using product rule) Chairs in an auditorium are to be labeled with a letter and a number between 1 and 100 - what is the largest number of chairs that can be labeled uniquely? Two tasks: –assign one of 26 letters –assign one of 100 integers By product rule, there are 26*100 different ways a chair can be labeled - so there are 2600 unique labels

9. 9 Example 4 (using product rule) How many bit strings of length 8 are there? There are 2 possible values for each bit So there are 2*2*2*2*2*2*2*2 or 2 8 (256) different bit strings

10. 10 Example 5 (using product rule) How many different social security numbers are possible? –A SSN has 9 digits –Each digit has one of 10 possible values –So there are 10 9, or 1 billion possible social security numbers

11. 11 Example 6 - counting functions How many functions are there from a set with m elements to one with n elements? –The n elements in the codomain correspond to each of m elements in the domain –So there are n*n* … * n = n m functions from a set with m elements to one with n elements

12. 12 Example 7 - counting one-to-one functions How many one-to-one functions are there from a set with m elements to one with n elements? –If m>n there are no one-to-one functions - so let m <= n –Suppose elements in domain are a 1, a 2, … a m ; there are n ways to choose value of function at a 1 –Since the function is one to one, the number of choices for a 2 are n-1

13. 13 Example 7 In general, the value at a k can be chosen in n-(k+1) ways By the product rule, there are: n(n-1)(n-2)*…*(n-m+1) one-to-one functions from a set of m elements to a set of n elements

14. 14 Example 8 - assigning telephone numbers Phone numbers in North America are specified by a numbering plan: –10 digits, with 3-digit area code, 3-digit office code, and 4-digit station code –Some of these sequences are restricted to digits 1-9 or 0-1, while others may be 0-9 –If x=0..9, y=0..1 and n=2..9 the plan in place until the 1960s was nyx-nnx-xxxx the current plan is: nxx-nxx-xxxx

15. 15 Example 8 How many unique phone numbers are possible under each plan? Old plan:n y x n n x x x x x 8*2*10*8*8*10*10*10*10*10 = 1,024,000,000 New plan:n x x n x x x x x x 8*10*10*8*10*10*10*10*10*10 = 6,400,000,000

16. 16 Product rule & nested loops k = 0; for (int i 1 =0; i 1 <n 1 ; i 1 ++) for (int i 2 =0; i 2 <n 2 ; i 2 ++). for (int i m =0; i m <n m ; i m ++) k++; What is the value of k after the following code is executed? By the product rule, value of k is: n 1 *n 2 * … *n m

17. 17 Product rule & sets If A 1, A 2, … A m are finite sets then the number of elements in the Cartesian product of the sets is the product of the number of elements in each set: |A 1 x A 2 x … x A m | = |A 1 | x |A 2 | x … x |A m |

18. 18 Example 9: Combining sum rule and product rule Suppose each user on a computer system has a 6-8 character password, according to the following rules: –each character is an uppercase letter or digit –there must be at least 1 digit per password How many possible unique passwords are there?

19. 19 Example 9 Let P = total number of passwords: –P 6 : total number with 6 characters –P 7 : total number with 7 characters –P 8 : total number with 8 characters By the sum rule, P = P 6 +P 7 +P 8

20. 20 Example 9 To find P 6, P 7 and P 8 : –Find the number of strings of letters and digits that are n characters (including those with no digits) –Subtract from this the number of n-character strings with no digits –Result is number of n-character strings with at least one digit

21. 21 Example 9 By the product rule, the number of strings with 6 characters is: (10 + 26) 6 : –10 possible digits –26 possible letters –6 positions in the string The number of strings with no digits is 26 6 So the number of 6-character strings with at least one digit is: P 6 = 36 6 - 26 6 = 1,867,866,560

22. 22 Example 9 By the same logic, –P 7 = 36 7 - 26 7 = 70,332,353,920 –P 8 = 36 8 - 26 8 = 208,827,064,567 And P, which is P 6 + P 7 + P 8 = 2,684,483,063,360

23. 23 Example 10 How many strings of 3 decimal digits do not contain the same digit 3 times? Let D=total number of 3-digit strings –product rule says D=10*10*10=1000 –Let D 0 = total number of 3-digit strings containing just 0’s - there is one of these –Similarly, D 1 =1, D 2 =1, D 3 =1, etc. By the sum rule, the number of 3-digit strings not containing the same digit 3 times is: D - (D 0 +D 1 +…+D 9 ) = 1000-10 = 990

24. 24 Example 11 How many strings of 3 decimal digits end with an even digit? Using the product rule, where each “D” below represents the number of possible values for one digit of the string: D 1 * D 2 * D 3 = 10 * 10 * 5 = 500

25. 25 Example 12 How many strings of 3 digits contain exactly 2 digits that are 4s? There are 3 ways this could occur: x44, 4x4, or 44x The value of x could be any of the other 9 digits By the product rule, the number of possible strings is 3 * 9, or 27

26. 26 Principle of Inclusion/Exclusion If two tasks can be done at the same time, can’t simply use the sum rule to count the number of ways to do the tasks, because the ways to do both tasks at once will be counted twice To find the correct value, add the number of ways each task can be done, then subtract the number of ways to do both at once

27. 27 Example 13 How many 8-bit strings either start with a 1 or end with 2 0s? –By the product rule, there are 2 7 (128) ways to form an 8-bit string that starts with 1 –By the same logic, there are 2 6 (64) ways to form an 8-bit string that ends with 2 0s –Finally, there are 2 5 (32) ways to form an 8-bit string that starts with 1 and ends with 2 0s –So, to form a string that starts with 1 or ends with 2 0s (but not both), there are 128+64-32 = 160 possible ways

28. 28 Sets and the inclusion/exclusion principle The number of elements in the union of two sets is the sum of the number of elements in both sets minus the number of elements in the intersection of the sets: |A 1  A 2 | = |A 1 | + |A 2 | - |A 1  A 2 |

29. 29 Tree diagram Graphical representation that can be used to help solve a counting problem For example, the diagram on the next slide can be used to determine the number of bit strings with length 4 that do not have 2 consecutive 1s

30. 30 Tree diagram 0 1 01 10 0 1 00 101 0 We start with the two possibilities for the first digit (0 or 1) If the first digit is 1, the second must be 0 On the other hand, if the first is 0, the next could be either 0 or 1 For the third digit, there are 6 possible paths The number of possibilities for the 4th digit gives us the overall result - there are 8 different ways to form a 4-digit bit string that does not contain consecutive 1s

31. 31 Section 4.1 Basics of Counting -ends-