Counting, pigeonhole, permuntation, Permutations and Combination ,Binomial Theorems

FM/2004/Melikyan

Counting Pigeonhole Principle

Permutation Permutations and Combinations

The Binomial Theorem

Discrete Math presentation

Nazia NishatLecturer

Department of Software Engineering Daffodil International University

2

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:1. Product Rule2. Sum Rule

Product Rule

Let, we have four students.A B C D

we 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 arranging all the members of

a set into some sequence or order.Examples:

All permutations made with the letters a, b, c taking 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 A combination is a way of selecting items from a collection, such that (unlike permutations) 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 players

7! / (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 Theorem

The binomial expansions

reveal a pattern.

0

1

2 2 2

3 3 2 2 3

4 4 3 2 2 3 4

5 5 4 3 2 2 3 4 5

( ) 1

( )

( ) 2

( ) 3 3

( ) 4 6 4

( ) 5 10 10 5

x y

x y x y

x y x xy y

x y x x y xy y

x y x x y x y xy y

x y x x y x y x y xy y

8.4 A Binomial Expansion Pattern

• The 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 Pascal’s triangle.

8.4 A Binomial Expansion Pattern

Pascal’s Triangle

Row1 0

1 1 11 2 1 2

1 3 3 1 31 4 6 4 1 4

1 5 10 10 5 1 5

8.4 Binomial Coefficients

Binomial Coefficient

For nonnegative integers n and r, with r < n,

!!( )!n r

n nCr r n r

8.4 Binomial Coefficients

• The symbols and for the binomial

coefficients are read “n choose r”

• The values of are the values in the nth row

of Pascal’s triangle. So is the first number

in the third row and is the third.

n rCnr

nr

30

32

. 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

FACTORIAL

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

FM/2004/Melikyan

Definition 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:

1) 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,

PERMUTATIONS

A PERMUTATION of a set of distinct objects is anArrangement of the objects in a specific order,without repetitions.

Pn,n = n(n - 1)·...·3·2·1 = n! (n factors)

Generalization

Find 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

( , ) ( 1)( 2)...( 1)( , )! ( 1)( 2)...1

P n r n n n n rC n rr r r r

Examples

Find C(8,5)

Solution: C(8,5) =

2. Find C(8,8)

Solution: C(8,8) =

(8,5) 8(7)(6)(5)(4) 8(7)(6) 8(7) 565! 5(4)(3)(2)(1) 3(2)(1)

P

(8,8) 8(7)(6)(5)(4)(3)(2)(1) 18! 8(7)(6)(5)(4)(3)(2)(1)

P