Counting - Discrete Structures - Lecture Slides, Slides of Discrete Structures and Graph Theory

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One, two, three, we’re…

• Counting

Basic Counting Principles

• Counting problems are of the following kind:
• “ How many different 8-letter passwords are there?”
• “ How many possible ways are there to pick 11 soccer players out of a 20-player team?”
• Most importantly, counting is the basis for computing probabilities of discrete events.
• (“What is the probability of winning the lottery?”)

Basic Counting Principles

• Generalized sum rule:
• If we have tasks T 1 , T 2 , …, Tm that can be done in n 1 , n 2 , …, nm ways, respectively, and no two of these tasks can be done at the same time, then there are n 1 + n 2 + … + nm ways to do one of these tasks.

Basic Counting Principles

• The product rule:
• Suppose that a procedure can be broken down into two successive tasks. If there are n 1 ways to do the first task and n 2 ways to do the second task after the first task has been done, then there are n 1 n 2 ways to do the procedure.

Basic Counting Principles

• Generalized product rule:
• If we have a procedure consisting of sequential tasks T 1 , T 2 , …, Tm that can be done in n 1 , n 2 , …, nm ways, respectively, then there are n 1  n 2  …  nm ways to carry out the procedure.

Basic Counting Principles

• The sum and product rules can also be phrased in terms of set theory.
• Sum rule: Let A 1 , A 2 , …, Am be disjoint sets. Then the number of ways to choose any element from one of these sets is |A 1  A 2  …  Am | = |A 1 | + |A 2 | + … + |Am|.
• Product rule: Let A 1 , A 2 , …, Am be finite sets. Then the number of ways to choose one element from each set in the order A 1 , A 2 , …, Am is |A 1  A 2  …  Am | = |A 1 |  |A 2 |  …  |Am|.

Inclusion-Exclusion

• Task 2: Construct a string of length 8 that ends with 00.
• There are two ways to pick the first bit (0 or 1),
• two ways to pick the second bit (0 or 1), -. -. -.
• two ways to pick the sixth bit (0 or 1),
• one way to pick the seventh bit (0), and
• one way to pick the eighth bit (0).
• Product rule: Task 2 can be done in 2^6 = 64 ways.

Inclusion-Exclusion

• Since there are 128 ways to do Task 1 and 64 ways to do Task 2, does this mean that there are 192 bit strings either starting with 1 or ending with 00?
• No, because here Task 1 and Task 2 can be done at the same time.
• When we carry out Task 1 and create strings starting with 1, some of these strings end with 00.
• Therefore, we sometimes do Tasks 1 and 2 at the same time, so the sum rule does not apply.

Inclusion-Exclusion

• Since there are 128 ways to complete Task 1 and 64 ways to complete Task 2, and in 32 of these cases Tasks 1 and 2 are completed at the same time, there are
• 128 + 64 – 32 = 160 ways to do either task.
• In set theory, this corresponds to sets A 1 and A 2 that are not disjoint. Then we have:
• |A 1  A 2 | = |A 1 | + |A 2 | - |A 1  A 2 |
• This is called the principle of inclusion-exclusion.

Tree Diagrams

• How many bit strings of length four do not have two consecutive 1s?
• Task 1 Task 2 Task 3 Task 4 (1st^ bit) (2nd^ bit) (3rd^ bit) (4th^ bit)

0

0

(^0 )

1 1 0 (^1 0 ) 1 (^1 )

(^0 ) 1 1 0 There are 8 strings.

The 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 of the objects.
• Example 1: In our 60-student class, at least 12 students will get the same letter grade (A, B, C, D, or F).

The Pigeonhole Principle

• Example 2: Assume you have a drawer containing a random distribution of a dozen brown socks and a dozen black socks. It is dark, so how many socks do you have to pick to be sure that among them there is a matching pair?
• There are two types of socks, so if you pick at least 3 socks, there must be either at least two brown socks or at least two black socks.
• Generalized pigeonhole principle: 3/2 = 2.

Permutations and Combinations

• So how can we compute how many different subsets of people can be picked (that is, we want to disregard the order of picking)?
• To find out about this, we need to look at permutations.
• A permutation of a set of distinct objects is an ordered arrangement of these objects.
• An ordered arrangement of r elements of a set is called an r-permutation.

Permutations and Combinations

• Example: Let S = {1, 2, 3}.
• The arrangement 3, 1, 2 is a permutation of S.
• The arrangement 3, 2 is a 2-permutation of S.
• The number of r-permutations of a set with n distinct elements is denoted by P(n, r).
• We can calculate P(n, r) with the product rule:
• P(n, r) = n(n – 1)(n – 2) …(n – r + 1).
• (n choices for the first element, (n – 1) for the second one, (n – 2) for the third one…)