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Section 4.1

Basics of Counting

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Basic Counting Principles:Sum Rule

• Suppose you have two tasks to perform:– The first task can be done in n1 ways;

– The second task can be done in n2 ways;– If the two tasks can’t be done simultaneously, then

there are n1 + n2 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

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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

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Example 2 (using sum rule)

What is the value of k after the following code is executed?

k = 0;for (int i1=0; i1<n1; i1++)

k++;for (int i2=0; i2<n2; i2++)

k++;...for (int im=0; im<nm; im++)

k++;

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Example 2

• Initial value of k is 0• There are m loops• Each iteration adds 1 to k• Let Ti = task of traversing the ith loop; can be

done ni ways, since traversal is done ni times• Since no 2 tasks can be done at once, final value

of k, which is the number of ways to do Ti, where i = 1 … m is:

n1 + n2 + … + nm

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Sum rule & sets

• If A1, A2, … , Am are disjoint sets:

A1 A2 … Am =

• Then the number of elements in the union of these sets = the sum of the number of elements in each set:

|A1 A2 … Am| = |A1| + |A2| + … + |Am|

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Basic Counting Principles: Product Rule

• Suppose a procedure can be broken down into 2 tasks, with n1 ways to do the first task and n2 ways to do the second task

• After the first task is done, there are n1n2 ways to the do the procedure

• Can extend product rule as sum rule was extended - in a procedure consisting of tasks T1, T2, … Tm, each of which can be done n ways, there are n1*n2* … *nm ways to carry out the procedure

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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

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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 28 (256) different bit strings

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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 109, or 1 billion possible social

security numbers

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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 = nm functions from a

set with m elements to one with n elements

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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 a1, a2, … am; there are n ways to choose value of function at a1

– Since the function is one to one, the number of choices for a2 are n-1

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Example 7

• In general, the value at ak 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

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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

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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

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Product rule & nested loops

k = 0;for (int i1=0; i1<n1; i1++)

for (int i2=0; i2<n2; i2++)...for (int im=0; im<nm; im++)

k++;

What is the value of k after the following code is executed?

By the product rule,value of k is:n1*n2* … *nm

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Product rule & sets

• If A1, A2, … Am 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:

|A1 x A2 x … x Am| = |A1| x |A2| x … x |Am|

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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?

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Example 9

• Let P = total number of passwords:– P6: total number with 6 characters

– P7: total number with 7 characters

– P8: total number with 8 characters

• By the sum rule, P = P6+P7+P8

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Example 9

• To find P6, P7 and P8:

– 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

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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 266

• So the number of 6-character strings with at least one digit is: P6 = 366 - 266 = 1,867,866,560

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Example 9

• By the same logic,– P7 = 367 - 267 = 70,332,353,920

– P8 = 368 - 268 = 208,827,064,567

• And P, which is P6 + P7 + P8 = 2,684,483,063,360

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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 D0 = total number of 3-digit strings containing just 0’s - there is one of these

– Similarly, D1=1, D2=1, D3=1, etc.

• By the sum rule, the number of 3-digit strings not containing the same digit 3 times is: D - (D0+D1+…+D9) = 1000-10 = 990

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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:D1 * D2 * D3 = 10 * 10 * 5 = 500

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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

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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

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Example 13

• How many 8-bit strings either start with a 1 or end with 2 0s?– By the product rule, there are 27 (128) ways to form an 8-bit

string that starts with 1

– By the same logic, there are 26 (64) ways to form an 8-bit string that ends with 2 0s

– Finally, there are 25 (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

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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:

|A1 A2| = |A1| + |A2| - |A1 A2|

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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

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Tree diagram0

10 1

1 001

• 00 0

10 0

1 10

0

We start with thetwo possibilities for the first digit (0 or 1)If the first digit is 1, thesecond must be 0On the other hand,if the first is 0,the next could beeither 0 or 1For the third digit,there are 6 possiblepaths

The numberof possibilitiesfor the 4th digit gives usthe overall result - thereare 8 differentways to form a 4-digit bitstring that doesnot containconsecutive 1s

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Section 4.1

Basics of Counting

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