# counting, pigeonhole, permuntation, permutations and combination ,binomial theorems

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What is Counting?

FM/2004/Melikyan

CountingPigeonhole PrinciplePermutationPermutations and CombinationsThe Binomial Theorem

Discrete Math presentationNazia NishatLecturerDepartment of Software Engineering Daffodil International University2

About Counting:Counting is a important part of Discrete Mathematics to find out the number of all possible outcomes for a series of events.

Basic Principles of Counting:Product RuleSum Rule

Product Rule

Let, we have four students.A B C Dwe have to find best two students among them. So we have two works to done.The ways for completing task one is K1The ways for completing task two is K2Now, K1 = 4 [because, we have 4 option] K2 = 3 [because, we have 3 option]So, the totals ways to complete the task is (4*3) or 12.

Sum RuleSuppose, we have to write an article and we have two list of topics.First list contains 6 topics.Second list contains 8 topic.

Total number of possible option of topics are 6+8 or 14.We can choose any one option of them.

Pigeonhole Principle

The Pigeonhole Principle states that if 2 or more pigeons are placed in holes, then one hole must contain two or more pigeons . Also Called box principle .n = Number of hole . K = Total Item .

if (K / n >= 2 (Ceiling)) What is Pigeonhole Principle ? Condition :

Example for Pigeonhole Principle

n = 3 K = 7

Ans = 7 / 3 = 3 ( Ceiling )

At least 1 bike Carry 3 or more passages

Where he go ??At least 1 bike Carry 3 or more passages

Permutation Permutation relates to the act of arrangingall the members of asetinto somesequenceororder.Examples: All permutations made with the lettersa,b,ctaking all at a time are:(abc,acb,bac,bca,cab,cba).Permutation Formula:nPr = n(n- 1)(n- 2) ... (n-r+ 1) = n!/(n-r)! How many ways can we award 3 Student from 7 Student ?7!/(7-3)! = 7!/4! = 7*6*5*4*3*2*1/4*3*2*1 = 7*6*5 = 210

Combination Acombinationis a way of selecting items from a collection, such that (unlikepermutations) the order of selection does not matter.Examples: Various groups of 2 out of four persons A, B, C, D are:AB, AC, AD, BC, BD, CD.

Pormula : nCr= n! / (r!)(n- r)!

Make a team of 3 players out of 7 players7! / (3!)(7!-3!)= 7! / (3!)(4!)= 7*6*5*4*3*2*1 / (3*2*1)(4*3*2*1)= 7*6*5 / 3*2*2

Permutations and Combinations

Combination: Picking a team of 3 people from a group of 10. 10C3 = 10!/(7! 3!) = 10 9 8 / (3 2 1) = 120. Permutation: Picking a President, VP and Waterboy from a group of 10. 10P3 = 10!/7! = 10 9 8 = 720.

Combination: Choosing 3 desserts from a menu of 10. 10C3 = 120. Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. 10P3 = 720.

8.4 The Binomial TheoremThe binomial expansions

reveal a pattern.

8.4 A Binomial Expansion PatternThe expansion of (x + y)n begins with x n and ends with y n .The variables in the terms after x n follow the pattern x n-1y , x n-2y2 , x n-3y3 and so on to y n . With each term the exponent on x decreases by 1 and the exponent on y increases by 1.In each term, the sum of the exponents on x and y is always n.The coefficients of the expansion follow Pascals triangle.

8.4 A Binomial Expansion PatternPascals Triangle Row

8.4 Binomial CoefficientsBinomial CoefficientFor nonnegative integers n and r, with r < n,

8.4 Binomial CoefficientsThe symbols and for the binomial coefficients are read n choose r

The values of are the values in the nth rowof Pascals triangle. So is the first number

in the third row and is the third.

. For more complicated problems, we will need to develop two important concepts: permutations and combinations. Both of these concepts involve what is called the factorial of a number. Permutations and Combinations

19The slides for this text are organized into chapters. This lecture covers Chapter 1.

Chapter 1: Introduction to Database SystemsChapter 2: The Entity-Relationship ModelChapter 3: The Relational ModelChapter 4 (Part A): Relational AlgebraChapter 4 (Part B): Relational CalculusChapter 5: SQL: Queries, Programming, TriggersChapter 6: Query-by-Example (QBE)Chapter 7: Storing Data: Disks and FilesChapter 8: File Organizations and IndexingChapter 9: Tree-Structured IndexingChapter 10: Hash-Based IndexingChapter 11: External SortingChapter 12 (Part A): Evaluation of Relational OperatorsChapter 12 (Part B): Evaluation of Relational Operators: Other TechniquesChapter 13: Introduction to Query OptimizationChapter 14: A Typical Relational OptimizerChapter 15: Schema Refinement and Normal FormsChapter 16 (Part A): Physical Database DesignChapter 16 (Part B): Database TuningChapter 17: SecurityChapter 18: Transaction Management OverviewChapter 19: Concurrency ControlChapter 20: Crash RecoveryChapter 21: Parallel and Distributed DatabasesChapter 22: Internet DatabasesChapter 23: Decision SupportChapter 24: Data MiningChapter 25: Object-Database SystemsChapter 26: Spatial Data ManagementChapter 27: Deductive DatabasesChapter 28: Additional Topics

FACTORIALFor n a natural number, n! = n(n - 1)(n - 2)...321 0! = 1 n! = n(n - 1)!1! = 12! = 23! = 64!= 3!*4 = 24

FM/2004/MelikyanDefinition of n factorial (!)n! = n(n-1)(n-2)(n-3)1 How it is used in counting:

Example. Solution: Let A,B,C symbolize the 3 number. They must fill 3 slots ___ ___ ___ ___ .

Two problems illustrating combinations and permutations. Consider the following two problems:

Consider the set { p , e , n} How many two-letter words (including nonsense words) can be formed from the members of this set?

We will list all possibilities: pe, pn, en, ep, np, ne , a total of 6.

2) Consider the six permutations of { p, e, n} which are grouped in three pairs of 2. Each pair corresponds to one combination of 2. pe, ep, np, pn, en, ne,

PERMUTATIONSA PERMUTATION of a set of distinct objects is anArrangement of the objects in a specific order,without repetitions. Pn,n = n(n - 1)...321 = n! (n factors)

GeneralizationFind P(5,5) , the number of arrangements of 5 objects taken 5 at a time.

Answer: P(5,5) = 5(5-1)(5-5+1) = 5(4)(3)(2)(1)=120.

Application: A bookshelf has space for exactly 5 books. How many different ways can 5 books be arranged on this bookshelf?

combination

Examples Find C(8,5)

Solution: C(8,5) =

2. Find C(8,8)

Solution: C(8,8) =

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