2.Derangements A derangement of {1,2,…,n} is a permutation i1i2…in of {1,2,…,n} in which no integer is in its natural position: i11,i22,…,inn. We denote.

Slides:



Advertisements
Similar presentations
THE WELL ORDERING PROPERTY Definition: Let B be a set of integers. An integer m is called a least element of B if m is an element of B, and for every x.
Advertisements

7.6 Applications of Inclusion/Exclusion Problem: Find the number of elements in a set that have none of n properties P1, P2,.. Pn.
CN College Algebra Ch. 11: Sequences 11.3: Geometric Sequences Goals: Determine if a sequence is geometric. Find a formula for a geometric sequence. Find.
Chapter 8 The principle of Inclusion and Exclusion Yen-Liang Chen Dept of Information Management National Central University.
1 Section 3.3 Mathematical Induction. 2 Technique used extensively to prove results about large variety of discrete objects Can only be used to prove.
4.4.2 Combinations of multisets
Sequences and Series (T) Students will know the form of an Arithmetic sequence.  Arithmetic Sequence: There exists a common difference (d) between each.
7.4 Generating Functions Definition 1: The generation function for the sequence a 0, a 1,...,a k,... of real numbers is the infinite series G(x) = a 0.
Chapter 6 The Inclusion-Exclusion Principle and Applications
Chapter 8. Section 8. 1 Section Summary Introduction Modeling with Recurrence Relations Fibonacci Numbers The Tower of Hanoi Counting Problems Algorithms.
4.6.2 Exponential generating functions
Lesson 4-1 Polynomial Functions.
Lecture#16 Discrete Mathematics. Recursion Now, 1 is an odd positive integer by the definition base. With k = 1, = 3, so 3 is an odd positive integer.
1 Chapter 7 Generating functions. 2 Summary Generating functions Recurrences and generating functions A geometry example Exponential generating functions.
Chapter 3 Permutations and combinations
Chapter 2 Mathematical preliminaries 2.1 Set, Relation and Functions 2.2 Proof Methods 2.3 Logarithms 2.4 Floor and Ceiling Functions 2.5 Factorial and.
4.5.2 Applications of Inclusion-Exclusion principle
Chapter 7 Advance Counting Techniques. Content Recurrence relations Generating function The principle of inclusion-exclusion.
The Pigeonhole Principle. The pigeonhole principle Suppose a flock of pigeons fly into a set of pigeonholes to roost If there are more pigeons than pigeonholes,
4.6 Generating functions Generating functions
Mathematical Induction Section 5.1. Climbing an Infinite Ladder Suppose we have an infinite ladder: 1.We can reach the first rung of the ladder. 2.If.
Lecture # 20 Sequence & Series
Chapter 5 1. Chapter Summary  Mathematical Induction  Strong Induction  Recursive Definitions  Structural Induction  Recursive Algorithms.
CMSC Discrete Structures
Section 6.2 Calculating Coefficients Of Generating Functions
Permutations and Combinations
Relations, Functions, and Matrices
Copyright © Cengage Learning. All rights reserved.
Modeling with Recurrence Relations
Chapter 4: Linear Differential Equations
Permutations and Combinations
Calculating Coefficients Of Generating Functions
To Exclude Or Not To Exclude?
Induction and recursion
A power series with center c is an infinite series where x is a variable. For example, is a power series with center c = 2.
Discrete Mathematics Lecture#14.
The Rational Zero Theorem
CS 2210 Discrete Structures Counting
General Logarithmic and Exponential Functions
4.4 Real Zeros of Polynomial Functions
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
CSE15 Discrete Mathematics 02/27/17
4.6.2 Exponential generating functions
Basic Counting.
182A – Engineering Mathematics
Other Forms of Induction
Copyright © Cengage Learning. All rights reserved.
The Fundamental Theorem of Algebra And Zeros of Polynomials
Copyright © Cengage Learning. All rights reserved.
Counting, Permutations, & Combinations
5.1 Power Series Method Section 5.1 p1.
The Rational Zero Theorem
Rational Root Theorem.
CMSC Discrete Structures
Solving Recurrence Relations
Sequences and the Binomial Theorem
Copyright © Cengage Learning. All rights reserved.
4.7 Recurrence Relations P13, P100
Applications of Inclusion-Exclusion: Selected Exercises
One-to-One and Onto, Inverse Functions
Advanced Analysis of Algorithms
CMSC Discrete Structures
4.4 Permutations and Combinations of multisets
Recurrence Relations Discrete Structures.
Surjective functions from N to X, up to a permutation of N
Mathematical Induction
Chapter 7 Advanced Counting Techniques
More Nonlinear Functions and Equations
Chapter 2 Limits and the Derivative
Power Series Solutions of Linear DEs
Presentation transcript:

2.Derangements A derangement of {1,2,…,n} is a permutation i1i2…in of {1,2,…,n} in which no integer is in its natural position: i11,i22,…,inn. We denote by Dn the number of derangements of {1,2,…,n}. Theorem 4.15:For n1,

Proof: Let S={1,2,…,n} and X be the set of all permutations of S Proof: Let S={1,2,…,n} and X be the set of all permutations of S. Then |X|=n!. For j=1,2,…,n, let pj be the property that in a permutation, j is in its natural position. Thus the permutation i1,i2,…,in of S has property pj provided ij=j. A permutation of S is a derangement if and only if it has none of the properties p1,p2,…,pn. Let Aj denote the set of permutations of S with property pj ( j=1,2,…,n).

Example:(1)Determine the number of permutations of {1,2,3,4,5,6,7,8,9} in which no odd integer is in its natural position and all even integers are in their natural position. (2) Determine the number of permutations of {1,2,3,4,5,6,7,8,9} in which four integers are in their natural position.

3. Permutations with relative forbidden position A Permutations of {1,2,…,n} with relative forbidden position is a permutation in which none of the patterns i,i+1(i=1,2,…,n) occurs. We denote by Qn the number of the permutations of {1,2,…,n} with relative forbidden position. Theorem 4.16:For n1, Qn=n!-C(n-1,1)(n-1)!+C(n-1,2)(n-2)!-…+(-1)n-1 C(n-1,n-1)1!

Proof: Let S={1,2,…,n} and X be the set of all permutations of S Proof: Let S={1,2,…,n} and X be the set of all permutations of S. Then |X|=n!. j(j+1), pj(1,2,…,n-1) Aj: pj Qn=Dn+Dn-1

4.6 Generating functions 4.6.1 Generating functions Let S={n1•a1,n2•a2,…,nk•ak}, and n=n1+n2+…+nk=|S|,then the number N of r-combinations of S equals (1)0 when r>n (2)1 when r=n (3) N=C(k+r-1,r) when ni r for each i=1,2,…,n. (4)If r<n, and there is, in general, no simple formula for the number of r-combinations of S. A solution can be obtained by the inclusion-exclusion principle and technique of generating functions. 6-combination a1a1a3a3a3a4

xi1xi2…xik= xi1+i2+…+ik=xr r-combination of S Definition 1: The generating function for the sequence a0,a1,…,an,… of real numbers is the infinite series f(x)=a0+a1x+a2x2+…+anxn+…, and if only if ai=bi for all i=0,1, …n, …

We can define generating function for finite sequences of real numbers by extending a finite sequences a0,a1,…,an into an infinite sequence by setting an+1=0, an+2=0, and so on. The generating function f(x) of this infinite sequence {an} is a polynomial of degree n since no terms of the form ajxj, with j>n occur, that is f(x)=a0+a1x+a2x2+…+anxn.

Example: (1)Determine the number of ways in which postage of r cents can be pasted on an envelope using 1 1-cent,1 2-cent, 1 4-cent, 1 8-cent and 1 16-cent stamps. (2)Determine the number of ways in which postage of r cents can be pasted on an envelope using 2 1-cent, 3 2-cent and 2 5-cent stamps. Assume that the order the stamps are pasted on does not matter. Let ar be the number of ways in which postage of r cents. Then the generating function f(x) of this sequence {ar} is (1)f(x)=(1+x)(1+x2)(1+x4)(1+x8)(1+x16) (2)f(x)=(1+x+x2)(1+x2+(x2)2+(x2)3)(1+x5+(x5)2)) =1+x+2x2+x3+2x4+2x5+3x6+3x7+2x8+2x9+2x10+3x11 +3x12+2x13+ 2x14+x15+2x16+x17+x18。

Example: Use generating functions to determine the number of r-combinations of multiset S={·a1,·a2,…, ·ak }. Solution: Let br be the number of r-combinations of multiset S. And let generating functions of {br} be f(y), (1+y+y2+…)k=? f(y)

Example: Use generating functions to determine the number of r-combinations of multiset S={n1·a1,n2·a2,…,nk·ak}. Solution: Let generating functions of {br} be f(y), f(y)=(1+y+y2+…+yn1)(1+y+y2+…+yn2)…(1+y+y2+…+ynk) Example: Let S={·a1,·a2,…,·ak}. Determine the number of r-combinations of S so that each of the k types of objects occurs even times. f(y)=(1+y2+y4+…)k=1/(1-y2)k =1+ky2+C(k+1,2)y4+…+C(k+n-1,n)y2n+…

Example: Determine the number of 10-combinations of multiset S={3·a,4·b,5·c}. Solution: Let generating functions of {ar} be f(y), f(y)=(1+y+y2+y3)(1+y+y2+y3+y4)(1+y+y2+y3+y4+y5) =1+3y+6y2+10y3+14y4+17y5+18y6+17y7+14y8+10y9+6y10+3y11+y12

Example: What is the number of integral solutions of the equation x1+x2+x3=5 which satisfy 0x1,0x2,1x3? Let x3'=x3-1, x1+x2+x3'=4, where 0x1,0x2,0x3'

4.6.2 Exponential generating functions The number of r-combinations of multiset S={·a1,·a2,…,·ak} : C(r+k-1,r), generating function: The number of r-permutation of set S={a1,a2,…, ak} :p(n,r), generating function:

C(n,r)=p(n,r)/r! Definition 2: The exponential generating function for the sequence a0,a1,…,an,…of real numbers is the infinite series

Theorem 4.17: Let S be the multiset {n1·a1,n2·a2,…,nk·ak} where n1,n2,…,nk are non-negative integers. Let br be the number of r-permutations of S. Then the exponential generating function g(x) for the sequence b1, b2,…, bk,… is given by g(x)=gn1(x)·g n2(x)·…·gnk(x),where for i=1,2,…,k, gni(x)=1+x+x2/2!+…+xni/ni! . (1)The coefficient of xr/r! in gn1(x)·g n2(x)·…·gnk(x) is

Example: Let S={1·a1,1·a2,…,1·ak} Example: Let S={1·a1,1·a2,…,1·ak}. Determine the number r-permutations of S. Solution: Let pr be the number r-permutations of S, and

Example: Let S={·a1,·a2,…,·ak},Determine the number r-permutations of S. Solution: Let pr be the number r-permutations of S, gri(x)=(1+x+x2/2!+…+xr/r!+…),then g(x)=(1+x+x2/2!+…+xr/r!+…)k=(ex)k=ekx

Example:Let S={2·x1,3·x2},Determine the number 4-permutations of S. Solution: Let pr be the number r-permutations of S, g(x)=(1+x+x2/2!)(1+x+x2/2!+x3/3!) Note: pr is coefficient of xr/r!. Example:Let S={2·x1,3·x2,4·x3}. Determine the number of 4-permutations of S so that each of the 3 types of objects occurs even times. Solution: Let p4 be the number 4-permutations, g(x)=(1+x2/2!)(1+x2/2!)(1+x2/2!+x4/4!)

Example: Let S={·a1,·a2, ·a3},Determine the number of r-permutations of S so that a3 occurs even times and a2 occurs at least one time. Solution: Let pr be the number r-permutations of S so that a3 occurs even times and a2 occurs at least one time, g(x)=(1+x+x2/2!+…+xr/r!+…)(x+x2/2!+…+xr/r! +…) (1+x2/2!+x4/4!+…)=ex(ex-1)(ex+e-x)/2 =(e3x-e2x+ex-1)/2

Next: Recurrence Relations P13, P100

Exercise : 1.Determine the number of permutations of {1,2,3,4,5,6,7,8} in which no even integer is in its natural position. 2.Determine the number of permutations of {1,2,…,n} in which exactly k integers are in their natural positions. 3.Eight boys are seated around a carousel. In how many ways can they change seats so that each has a different boy in front of him? 4.Let S be the multiset {·e1,·e2,…, ·ek}. Determine the generating function for the sequence a0, a1, …,an, … where an is the number of n-combinations of S with the added restriction: 1) Each ei occurs an odd number of times. 2) the element e2 does not occur, and e1 occurs at most once. 5.Determine the generating function for the number an of nonnegative integral solutions of 2e1+5e2+e3+7e4=n 6.Determine the number of n digit numbers with all digits at least 4, such that 4 and 6 each occur an even number of times, and 5 and 7 each occur at least once, there being no restriction on the digits 8 and 9.