Chapter 7 Advance Counting Techniques

Content

Recurrence relations

Generating function

The principle of inclusion-exclusion

Recurrence relations

• Definition: a recurrence relation for the sequence {an} is an equation that express an in terms of the sequence , namely a1 , a2, …, an-1 , for all integers n with n>n0. where n0 is a nonnegative integer. A sequence is called a solution of a recurrence relation if its term satisfy the recurrence relation.

• The initial conditions for a sequence specify the terms that precede the first term where the recurrence relation takes effect.

• Example1: let {an} be a sequence that satisfies the recurrence relation an= an-1- an-2 for n=2,3,4,… and suppose that a0=3 and a1=5. what are a2 and a3?

• Example2: determine whether the sequence {an} is a solution of the recurrence relation

an= 2an-1- an-2 for n=2,3,4,….

(1) an= 3n

(2) an= 2n

(3) an= 5

Modeling with recurrence relation• Example3: Compound Interest

Suppose that a person depots $10000 in a saving account at a bank yielding 11% per year with interest compound annually. How much will be in the account after 30 years?

• Solution: Pn=Pn-1+0.11Pn-1=1.11Pn-1

using a iterative approach to find a formula Pn .

Pn=(1.11)nP0= (1.11)3010000=$228922.97

• Example4: Rabbits and the Fibonacci Numbers

A young pair of rabbits (one of each sex) is placed on an island. A pair of rabbits does not breed until they are 2 months old. After they are 2 months old, each pair of rabbits produces another pair each month. Find a recurrence relation for the number of pairs of rabbits on the island after n months, assuming that no rabbits ever die.

• Solution: fn=fn-1+fn-2 with f1=1 and f2=1.

• Example5: The Tower of Hanoi

The Hanoi Tower consists of three pegs mounted on a board together with disks of different sizes. Initially these disks are placed on the first peg in order of size, with the largest on the bottom. The rule of the puzzle allow disks to be moved one at a time from one peg to another as long as a disk is never placed on top of a smaller disk. The goal of the puzzle is to have all the disks on the second peg in order of size, with the largest on the bottom. Let Hn denote the number of moves needed to solve the Tower of Hanoi problem with n disks. Set up a recurrence relation for the sequence {Hn}.

• Solution: begin with n disk on peg one, we can transfer the top n-1 disks following the rules of puzzle, to peg three using Hn-1 moves. Hence we have Hn=2 Hn-1+1 with H1=1.

• using the iterative approach to solve this recurrence relation.

1

22 2

2 33 3

1 2 31

1 2 3

2 1

2(2 1) 1 2 1

2 (2 1) 2 1 2 2 1

2 2 2 2 1

2 2 2 2 1

2 1

n n

n n

n n

n n n

n n n

n

H H

H H

H H

H

• Example6: Find a recurrence relation and give

initial conditions for the number of bit strings of

length n that do not have two consecutive 0s.

How many such bit strings are there of length

five?

• Solution: let an denote the number of bit strings of length n that do not have two consecutive 0s. To obtain a recurrence relation for {an}, note that by the sum rule, the number of bit strings of length n that do not have two consecutive 0s equals the number of such bit strings ending with a 0 plus the number of such bit strings ending with a 1.

1Any bit string of length n-1 with

no two consecutive 0sEnd with a 1 1na

Any bit string of length n-2 with no two consecutive 0s

End with a 0 1 0 2na

We will assume that , so that the bit string has at least 3 bits.

Hence we have an=an-1+an-2 for with a1=2.

3n

3n

• Example7: Codeword Enumeration

A computer system consider a string of decimal

digits a valid codeword if it contains an even

number of 0 digits. For instance, 1230407869 is

valid, whereas 120987045608 is not valid. Let

an be the number of valid n-digit codeword.

Find a recurrence relation for an.

• Solution: There are two ways to construct a valid string with n digits from a string with one fewer digit.

(1) A valid string of n-digits can be obtained by appending a valid string of n-1 digits with a digit other than 0. namely 9an-1 ways there are.

(2) A valid string of n-digits can be obtained by appending a 0 to a string of length n-1 that is not valid. Then there are 10n-1-an-1.

Hence an= 9an-1+ 10n-1-an-1= 8an-1+ 10n-1 with a1=9.

Solving Recurrence Relations-1• Definition1 : A linear homogeneous recurrence

relation of degree k with constant coefficients is a recurrence relation of the form

where are real numbers, and .

• Example1: (1) Pn=Pn-1+0.11Pn-1=1.11Pn-1

(2) fn=fn-1+fn-2 with f1=1 and f2=1.

(3) Hn=2 Hn-1+1 with H1=1

1 1 2 2n n n k n ka c a c a c a 1 2, , , kc c c 0kc

• Characteristic equation and characteristic roots

Definition2: assume that a recurrence relation is

then the equation

is called characteristic equation ; and the solution of the equation are called characteristic roots.

1 1 2 2n n n k n ka c a c a c a

1 21 2 1 0k k k

k kr c r c r c r c

• Theorem1: Let c1 and c2 be real numbers. Suppose that r2-c1r-c2=0 has two distinct roots r1 and r2. then the sequence {an} is a solution of the recurrence relation an=c1an-1+c2an-2 if and only if for n=0,1,2,… where are constants and are determined by the initial conditions.

1 1 2 2n n

na r r 1 2,

• Example2: what is the solution of the recurrence relation an=an-1+2an-2 with a0=2 and a1=7 ?

• Solution: the characteristic equation is r2=r+2,

then the solution of the equation is r=2 and r=-1.

hence an=b12n+b2(-1)n with a0=2 and a1=7. thus

we have b1= 3 and b2=-1. namely

an=3 . 2n-(-1)n

• Theorem2: Let c1 and c2 be real numbers with c20, Suppose that r2- c1r- c2=0 has only one root r0. then a sequence {an} is a solution of the recurrence relation an=c1an-1+c2an-2 if and only if for n=0,1,2,… where are constants and are determined by the initial conditions.

1 0 2 0n n

na r nr 1 2,

• Example3: what is the solution of the recurrence relation an=6an-1-9an-2 with a0=1 and a1=6?

• Solution: the root of r2- 6r+9=0 is r=3 , hence

an=b13n+b2 n3n with a0=1 and a1=6.

an=3n+n3n with a0=1 and a1=6.

• Theorem3: Let c1 , c2 ,…, ck be real numbers . Suppose that rk- c1rk-1 -…-r- ck=0 has k distinct root r1, r2, …, rk. Then a sequence {an} is a solution of the recurrence relation

an=c1an-1+c2an-2 +…+ ckan-k

if and only if an =b1r1n+ b2r2

n+ …+ bkrkn for

n=0,1,2,… , where b1, … , bk are constants and are determined by the initial conditions.

• Example4: Find the solution of the recurrence

relation an=6an-1-11an-2 +6an-3 with a0=2, a1=5

and a2=15?

• Solution: the root of r3- 6r2+11r-6=0 is r=1,2,3;

hence an =b1+ b22n+ b33n and b1=1, b2=-1, b3=2

• Theorem4: Let c1 , c2 ,…, ck be real numbers . Suppose that rk- c1rk-1 -…-r- ck=0 has t distinct root r1, r2, …, rt with multiplicities m1, m2,…, mt and m1+m2+…+mt =k. Then a sequence {an} is a solution of the recurrence relation

an=c1an-1+c2an-2 +…+ ckan-k

if and only if an =(b10+ b11 n+…+ b1m1-1 nm1-1 )r1n+ +

…+ (bt0+ bt1 n+…+ btmt-1 nmt-1 ) rtn for n=0,1,2,… ,

where bij are constants and are determined by the initial conditions.

• Example5: what is the solution of the recurrence relation an=-3an-1-3an-2 -an-3 with a0=1, a1=-2 and a2=-1?

• Solution: the root of r3+3r2+3r+1=0 is r=-1of multiplicity three;

hence an =(b1+ b2n+ b3n2 )(-1)n

and b1=1, b2=3, b3=-2

Linear Nonhomogeneous Recurrence Relations with Constant Coefficients

• Definition: the recurrence relation

an=c1an-1+c2an-2 +…+ ckan-k +F(n)

is called Linear Nonhomogeneous Recurrence Relations with Constant Coefficients.

• an=c1an-1+c2an-2 +…+ ckan-k is called the associated homogeneous recurrence relation.

• Theorem5: If {an (p)} is a particular solution of the

nonhomogeneous linear recurrence relation with constant coefficients

an=c1an-1+c2an-2 +…+ ckan-k +F(n) ,

Then every solution is of the form {an (p) + an

(h) } , where {an

(h) } is a solution of the recurrence relation of the associated homogeneous recurrence relation

an=c1an-1+c2an-2 +…+ ckan-k .

• Example6: what is the solution of the recurrence relation an=5an-1-6an-2 +7n

with a0=2, a1=5 ?

• Solution: the solution of an=5an-1-6an-2 is

an (h) =b13n+b22n , since F(n)=7n,

assume that an (p) =c 7n , where c is a constant.

Hence c 7n =5 c 7n-1 -6 c 7n-2 + 7n ， we have c =49/20, namely that an

(p) =(49/20) 7n . Then we can determine the coefficients b1 and b2 .

• Theorem6: If {an} satisfies the nonhomogeneous

linear recurrence relation with constant coefficients

an=c1an-1+c2an-2 +…+ ckan-k +F(n)

where c1, c2,…, ck are real numbers and

F(n) =(bt nt+ bt-1nt-1 +…+ b1n+b0) sn

where b0, b1, … , bt and s are real numbers.

(1) If s is not a characteristic root, then there is a particular solution of the form

an (p) =(pt nt+ pt-1nt-1 +…+ p1n+p0) sn

(2) If s is a characteristic root with multiplicity m, then there is a particular solution of the form

an (p) =nm (pt nt+ pt-1nt-1 +…+ p1n+p0) sn

• Example7: what is the solution of the linear nonhomogeneous relation an=6an-1-9an-2 +F(n)

(1) F(n)=n2 2n (2) F(n)=3n (3) F(n)=n3

(1) Solution:

• let an (p) =(p2

n2 +p1 n+ p0) 2n .Then

(p2 n2 +p1 n+ p0) 2n= 6(p2

(n-1)2 +p1 (n-1)+ p0) 2n-1- 9(p2

(n-2)2 +p1 (n-2)+ p0) 2n-2 + n2 2n

we have that p0= 192 p1=48 p2=4 .

• The characteristic root is r=3. hence an=(b1+b2n)3n is a solution of an=6an-1-9an-2 .

• Definition1: The generating

function for the sequence a0,

a1, …, ak, … of real numbers

is the infinite series

G(x)= a0 + a1 x+ …+ak xk+…

Solving Recurrence Relations-2

• Example1: the generating function for the sequences {ak} with ak=3, ak=2(k+1) and ak=2k

are

• Example2: what is the generating function for the sequence 1,1,1,1,1?

• Solution: G(x)= 1+ x+ x2+x3 +x4

0 0 03 , 2( 1) , (2 )k k k

k k kx k x x

Useful Facts About Power Series• Example3: the function

f(x)=1/(1-x)=1+ x+ x2+x3 +x4+ x5+… is the generating functions of the sequence 1,1,… the function f(x)=1/(1-ax)=1+ ax+ a2 x2+… is the generating

functions of the sequence 1,a, a2,… the function

f(x)=1/(1-x)2 =1+ 2x+ 3x2+4x3 +5x4+ 6x5+… is the generating functions of the sequence 1,2,3,

…, k+1, …

• Definition2: Let u be a real number and k a nonnegative integer. Then the extended binomial coefficient C(u,k) is defined by

• Exmaple4: C(-2,3)=-4 C(1/2,3)=1/16

• Exmaple5: when n<0 is an integer, then

( 1) ( 1) / ! 0( , )=

1 0

u u u k k if kC u k

if k

( )( 1) ( 1)!

( 1) ( 1) ( 1)!

1( 1) ( 1)!( 1)

!( 1)!

r

rr

n n n n rrr

n n n rr

n rn rr n r

• Threorem2: THE EXTENDED BINOMIAL THREOREM

Let x be a real number with -1<x<1 and u be a real number . Then

• Example6:

0

(1 )u k

k

ux x

k

0 0

1(1 ) ( 1)n k k k

k k

n n kx x x

k k

0

1(1 ) n k

k

n kx x

k

Counting Problems and Generating Functions

• Example7: Find the number of solutions of

e1+e2 +e3 =17, where e1, e2 and e3 are nonnegative integers with 2e1 5, 3e2 6 and 4e1 7.

• Solution: the number of solutions with the indicated constraints is the coefficient of x17 in the expansions of:

(x2 + x3 + x4 + x5) (x3 + x4 + x5 +x6) (x4 + x5 + x6 + x7 )

This follows that the coefficient of x17 is 3.

• Example8: In how many different ways can eight

identical cookies be distributed among three

distinct children if each child receives at least two

cookies and no more than four cookies.

• Solution: the coefficient of x8 in the product

(x2 + x3 + x4)3 is 7.

• Example9: use generating functions to determine the

number of ways to insert tokens worth $1, $2,and $5

into a vending machine to pay for an item that costs

r dollars in both the cases when the order in which

the tokens are inserted does not matter and when the

order does matter.

• Solution: case1: no order is considered

the answer is the coefficient of xr in the generating

function (1+x+x2 + …)(1 +x2 +x4+ …)(1+x5 +x10 + …)

• case2: when the order does matter, then the

number of ways to insert exactly n tokens to

produce a total of r dollars is the coefficient of

xr in the generating function (x+x2 + x5)n , since

each of the r tokens may be a $1 token, a $2

token, or a $5 token. Since any number tokens

may be inserted, the number of ways to

produce r dollars is the coefficient of xr in

1+ (x+x2 + x5)+ (x+x2 + x5)2 +…=1/1- (x+x2 + x5)

• Example10: use generating function to find the number of k-combinations of a set with n elements. Assume that the Binomial Theorem has been established.

• Solution: each of the n elements in the set contributes the term (1+x) to the generating function f(x)=(1+x)n , by the binomial theorem , we have

hence

0( )

n k

k

nf x x

k

!( , )

!( )!

n nC n k

k n kk

• Example11: use generating function to find the

number of r-combinations from a set with n

elements when repetition of elements is allowed.

• Solution: Let G(x)=(1+x+x2 +…)n, then the

coefficient of xr is the answer. Hence

G(x)=(1+x+x2 +…)n=(1+x)-n=C(n+r-1,r)xr

Using Generating Functions TO Solve Recurrence Relations

• Example12: solve the recurrence relation

an=6an-1-9an-2 with a0=1 and a1=6.

• Solution: let , Then

0( )= n

nnG x a x

1 22 2

2

0 0

2

( ) 1 6 (6 9 )

6 9 1

(6 9 ) ( ) 1

n nn n nn n

n nn nn n

G x a x x a a x

x a x x a x

x x G x

Using the recurrence relation, we have

2

2 0

0

1( )

1 6 921

( )( 3)(1 3 )

( 1)3

n n

n

n n

n

G xx x

xnx

n x

• Example13: Solve the recurrence relation

an=8an-1+10n-1 with a1=9.

• Solution: Let

then

0

( ) kk

k

G x a x

1

1

1 1

0 1

( ) 1 (8 10 )

8 10

8 ( )1 10

k kk

k

k k kk

k k

G x a x

x a x x x

xxG x

x

Hence we have

0

1 9( )

(1 8 )(1 10 )

1 1 1( )

2 1 8 1 101

(8 10 )2

n n n

k

xG x

x x

x x

x

＋

Inclusion-Exclusion

• The Principle of Inclusion-Exclusion

• Theorem1: Let A1, A2 , … , An be finite sets. Then

1 2

1 1

1

11 2( 1)

n

i i ji n i j n

i j ki j k n

nn

A A A

A A A

A A A

A A A

1 An Alternative Form of Inclusion-Exclusion

• Let Ai be the subset containing the elements that property Pi . The number of elements with all the properties Pi1, Pi2 , … , Pik will be denoted by N(Pi1Pi2 … Pik ). Writing these quantities in terms of sets, we have

If the number of elements with none of the

properties P1,P2 ,…, Pn , is denoted by and the number of elements

in the set is denoted by N.

1 2( )nN PP P

1 2 1 2( )

k ki i i i i iA A A N P P P

From the Principle of Inclusion-Exclusion, we see that

1 2 1 2

1 1 1

1 2

( )

( ) ( ) ( )

( 1) ( )

n n

i i j i j ki n i j n i j k n

nn

N PP P N A A A

N N P N PP N PP P

N PP P

• Example1: how many solutions does

x1+x2+x3=11 have, where x1 , x2 and x3 are nonnegative integers with x1 , x2 and x3

• Solution: let P1 denote the property of element of the set that x1and P2 : x2 , P3: x3

Then 1 2 3 1 2 3

1 2 1 3 2 3 1 2 3

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

N PP P N N P N P N P

N PP N PP N P P N PP P

• N=C(3+11-1,11)=78

• N(P1)=C(3+7-1,7)=36

• N(P2)=C(3+6-1,6)=28

• N(P3)=C(3+4-1,4)=15

• N(P1P2)=C(3+2-1,2)=6

• N(P1P3)=C(3+0-1,0)=1

• N(P2P3)=0

• N(P1P2 P3)=0

• =61 2 3( )N PP P

THE SIEVE OF ERATOSTHENES

• To find the number of primes not to exceeding a specified positive integer.

• For instance, to find the number of primes not to exceeding 100.

• Note that a composite integer not to exceeding 100 must have a prime factor not to exceeding 10.

• The primes not to exceeding 100 are divisible by none of 2,3,5 and 7.

• Let P1, P2, P3, P4 denote an integer is divisible by 2,3 5,7 respectively.

• Then the number of primes not to exceeding 100 is 1 2 3 44 ( )N PP P P

1 2 3 4 1 2 3 4

1 2 1 3 1 4 2 3 2 4

3 4 1 2 3 1 2 4 2 3 4

1 2 3 4

( ) 99 ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )

99 50 33 20 14 16 10 7 6 4 2

3 2 1 0 0 21

N PP P P N P N P N P N P

N PP N PP N PP N P P N P P

N P P N PP P N PP P N P P P

N PP P P

The Number of Onto Functions • Example2: how many onto functions are there from

a set with six elements to a set with three elements?

• Solution: assume that the elements in the codomain are b1, b2, b3. let P1,P2,P3 be the properties that b1, b2 and b3 are not in the range of the function, respectively. note that a function is onto if and only if it has none of the properties P1,P2 and P3. hence

1 2 3 1 2 3

1 2 1 3 2 3 1 2 3

6 6 6

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3 (3,1)2 (3,2)1 540

N PP P N N P N P N P

N PP N PP N P P N PP P

C C

• Theorem1: let m and n be positive integers

with mn. Then , there are

Onto functions from a set with m elements to a

set with n elements.

1

( ,1)( 1) ( ,2)( 2)

( 1) ( , 1)1

m m m

n m

n C n n C n n

C n n

Derangements

• Example3: The Hatcheck Problem

A new employee checks the hats of n people at a

restaurant, forgetting to put claim check

numbers on the hats. When customers return for

their hats, the checker gives them back hats at

random from the remaining hats. What is the

probability that no one receives the correct hat?

• Remark: the answer is the number of ways the

hats can be arranged so that there is no hat in its

original position divided by n!.

• Definition: a derangement is a permutation of

objects that no object in its original position.

• For example: 21453 is a derangement of 12345.

• Let Dn denote the number of derangement of n

objects.

• Theorem2: the number of derangements of a

set with n elements is

1 1 1 1!(1 ( 1) )

1! 2! 3! !n

nD nn

• Proof: let a permutation have property Pi if it

fixes element i. The number of derangements

is the number of permutations having none of

the properties Pi for i=1,2,…, n. This means

that

1 2( )n nD N PP P

• Using the principle of inclusion-exclusion, it

follows that

1 2 1 2

1 1 1

1 2

( )

! ( ) ( ) ( )

( 1) ( )

n n

i i j i j ki n i j n i j k n

nn

N PP P N A A A

n N P N PP N PP P

N PP P

• We can see that

• N(Pi)=(n-1)!

• N(Pi Pj)=(n-2)!

• ……

• N(Pi1 Pi2 … Pim )=(n-m)!

• And

• And in general,

1

1

( ) ( ,1)( 1)!

( ) ( , 2)( 2)!

ii n

i ji j n

N P C n n

N PP C n n

1 2

1 21

( ) ( , )( )!m

i i imi i i n

N P P P C n m n m

• Hence we have

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

! ! !! ( 1)! ( 2)! ( 1) ( )!

1!( 1)! 2!( 2)! 0! !

1 1 1![1 ( 1) ]

1! 2! !

nn

n

n

D n C n n C n n n n

n n nn n n n n

n n n

nn

n 2 3 4 5 6 7

Dn /n! 0.5000 0.33330.375

00.3667 0.3681

0.3678

Table1 the probability of a derangement