discrete math by r.s. chang, dept. csie, ndhu1 chapter 9

Discrete Math by R.S. Chang, Dept. CSIE, NDHU 1

Chapter 9


Chapter 99.1 Introductory Examples

Ex. 9.1 Find the number of integer solutions toc c c c c c1 2 3 1 2 312 4 2 2 5 where , , .

c x x x x x c xc x x x x x c xc x x x x c x

x f x c x c x c x

f x generating function

14 5 6 7 8

1

22 3 4 5 6

2

32 3 4 5

3

1 2 3

: ( ): ( ): ( )

( ) ( ) ( ),

The coefficient of in ( ) =which is 14, is the answer.

( ) is called a for the distributions.

12


Chapter 99.1 Introductory Examples

Ex. 9.3 How many integer solutions are there for the equation

c c c c c ic

x x x x

f x x x x

g x x x x xx

x

ii

1 2 3 4

25

2 25 4

2 25 26 4

44

25 0 1 4

1

11

11

, , ?,

.

( )( )

For each the possibility can be described by1 + + Then the answer is the coefficient of in the generating function:

( ) = + + or

( ) = + +

2 25


Chapter 99.2 Definitions and Examples: Calculational Techniques

Def. 9.1 Let be a sequence of real numbers. The

function ( ) = is called the

generating function for the given sequence.

Ex. 9.4 For any Z

so (1 + ) is the generating function for the sequence

0

0

+

a a a

f x a a x a x a x

n xn n

xn

xnn

x

xn n n n

n

ii

i

n n

n

, , ,

, ( )

, , , , , , , , .

1 2

1 22

0

210 1 2

0 1 20 0 0



Ex. 9.5 (a) For Z (1 -

So 1 - x is the generating function for , ,

b) If and | |< 1, 1 = ( - )( + + x

So 1-

is the generating function for 1,1,1, .

(c)

Consequently, 1

+ +

n+1

+ '2

n x x x x

x

n x x x x

xddx x

xx

ddx

x x x x x x

n n

n s

, ) ( )( ).

, , , .

( ).

( )( ) ( )( )

( )

1

1 13

22

2 3 2 3

1 1

111 1 0 0

1 1

11

11 1 1 1

11 1 2 3 4

( - ) is the generating function of 1,2,3, .

21 x



Ex. 9.5 (continued)

And ( - )

is the generating function of 0,1,2,3, .

(d) Continuing from (c),

Hence +

( - ) generates 1 and

( + )

( - ) generates 0 1

2

2 2

x

xddx

x

x

x

xddx

x x x x x xx

xx x

x

1

1

1

10 2 3 1 2 3 4

1

12 3

1

12 3

2

2 3

2 3 2 2 2 2 3

32 2

32 2

( ) ( )( ) .

, , , ,

, , , , .



Ex. 9.6 (a) generating function of 1,1,0,1,1, is 11 - x

The generating function for 1,1,1,3,1,1, is 11 - x

b) Find the generating function for 0,2,6,12,20,30,42,

Therefore, the generating function is

x

x

a a a aax x

x

x

x

x

x

2

3

02

12

22

32

42

3 2 3

2

0 0 0 2 1 1 6 2 2 12 3 320 4 4

1

1 1

2

1

(, , , ,

, .( )

( ) ( ) ( )



Extension of binomial coefficientWith , Z and > 0, we have

If R, we use as the definition

of For example, if Z we have -

+

+

n r n rnr

nr n r

n n n n rr

n n n n n rr

nr

nnr

n n n n rr

n n n rr

n rr

r

r

!!( )!

( )( ) ( )!

.

( )( ) ( )!

. ,

( )( )( ) ( )!

( ) ( )( ) ( )!

( )

1 2 1

1 2 1

1 2 1 1 1 1

11

. . And for any real n, define

n0

1



Ex. 9.7 For n Z

Ex. 9.8 Find the coefficient of x in (1 - 2x)

Ex. 9.10 Find the coefficient of x in f(x) = (x

The coefficient of x in

is -47

+

5 -7

15 2

7

, ( ) ( )

.

( ) ( ) ( ) , .

) .

( ) ( )( )

.

( )

1 11

75

2 17 5 1

532 14 784

11

1

1

0

0

5 5

3 4

2 2 4 8

4

4

xn r

rx

nr

x

x

f x x x x x

x

x

n r

r

r

r

r

( )1 1207



Ex. 9.11 In how many ways can we select, with repetitions allowed,r objects from n distinct objects?

For each of the distinct objects, the geometric series+ + represents the possible choices for the

object. Considering all of the objects, the generating functionsis ( ) = ( + + and the required answer is the

coefficient of in ( ). ( ) = ( - )

So the answer is + -

2

-

nx x x

nf x x x x

x f x f x xni

x

n ii

xn r

r

n

r n i

i

i

i

1

1

1

1 1

3

2 3

0

0

) ,

. .



Ex. 9.13 Verify that for all Z

Since (1 + ) by comparison of coefficients,

the coefficient of in ( + ) which is must equal the

coefficient of in and that is

+nn

nni

x x

x xnn

xn n

xnn

x

n nn

n nn

nn

i

n

n n

n n

n n

, .

[( ) ] ,

, ,

,

2

1

12

0 1

0 1 1

2

02 2

2

2

n nr

nn r0

. , With the result

follows.



Ex. 9.14 Determine the coefficient of in ( - )( - )

partial fraction decomposition:

1

( - )( - ) or

1 = ( - ) By comparing coefficients, = , = -1, and = -1. Hence,

1

( - )( - )

2

xx x

x x

Ax

Bx

C

xA x B x x C x

A B C

x x x x x x

x

82

2

2

2 2

1

3 2

3 2 3 2 22 2 3 3

1

3 2

13

12

1

2

13

11 3

12

11 2

14

.

( )( )( ) ( ).

( ) ( / )

( / )

1

1 2 2( ( / ))x



Ex. 9.15 Use generating functions to determine how many four-element subsets of S={1,2,3,...,15} contain no consecutiveintegers.Let { be one such subset with 1 Let

for 2 and Then with 0

and 2 Therefore, the answer is the coefficient of in ( ) = ( + + x

which is -58

1

=

2

a a a aa a a a c a c a a

i c a c c c

c c cx f x x x x x x

i i i

ii

1 2 3 41 2 3 4 1 1

5 41

51 5

2 3 414 2 2 3 3 6 5

8

15 1

4 15 14

1 1

1 495

, , , }. ,

, . ,

, , .) ( ) ( ) ,

( ) .



Ex. 9.16 ( ) =( - )

generates 0,1,2,

and ( ) = ( + )

( - ) generates 0

Then ( ) = ( ) ( ) = where

2

=

f x x

xa a a

g x x x

xb b b

h x f x g x c x

c a b a b a b a b a b a bi k i i k ki i k i

kk

kk k k k k k k

ik

ik

ik

11

11 2

2

2 0 1 2

32 2

0 1 2

00 1 1 2 2 2 2 1 1 0

20

2 20

20

( , , , )

, , , ( , , , )

,

( ) ( )

21

22

1 2 16

12

20

30

2

2

0 1 2 0 1 2

k i i kk k

kk k k

k k a a a b b b

ik

ik ( ) ( )( )

( ) . , , , , , , . Convolution of and


Chapter 99.3 Partitions of Integers

Partition a positive integer n into positive summands and seeking the number of such partitions, without regard to order.For example, p(1)=1: 1 p(2)=2: 2=1+1 p(3)=3: 3=2+1=1+1+1 p(4)=5: 4=3+1=2+2=2+1+1=1+1+1+1 p(5)=7: 5=4+1=3+2=3+1+1=2+2+1=2+1+1+1 =1+1+1+1+1

We should like to obtain p(n) for a given n without having tolist all the partitions. We need a tool to keep track of the numbers of 1's, 2's, ..., n's that are used as summands for n.



keep track of 1's: 1 + +keep track of 2's: 1 + +

keep track of ' s: 1 + +For example, p(10) is the coefficient of in

( ) = ( + +

In general, ( ) = generate the sequence ( ), (1),

2

4

2k

10

2

x x xx x x

k x x xx

f x x x x x x

x x x x x

P xx

p p

k k

ii

ii

3

2 6

3

2 4 10

2 3 10 1

10

1

1 1 11

11

1

1

1

1

1

1

11

10

)( ) ( )

.



Ex. 9.18 Find the generating function for the number of ways anadvertising agent can purchase n minutes of air time if time slotsfor commercials come in blocks of 30, 60, or 120 seconds.

Let 30 seconds represent one time unit. Then the answer is thenumber of integer solutions to the equation + + =with 0 , , . The associated generating function is

( ) = ( + +

and the coefficient of is the answer.

2

a b c na b c

f x x x x x x x

x x xx n

2 4 2

1 1 11

11

1

1

1

2 4 4 8

2 42

)( )( )



Ex. 9.19 Find the generating function for pd(n), the number ofpartitions of a positive integer n into distinct summands.For any Z either is not used as a summand or it is. This can be accounted for by the polynomial 1 +Consequently, the generating function is

the generating function of partitioning into odd summands

+k kx

P x x x x x

xx

x

x

x

x

x

x x xx x x x x x P x

k

di

i

o

,.

( ) ( )( )( ) ( )

( )( )( ) ( )

1 1 1 1

11

1

1

1

1

1

1

11

1

11 1 1

2 3

12 4

2

6

3

8

4 32 3 6 5 10



Ex. 9.21 Partition into odd summands but each such odd summandsmust occur an odd number of times-or not at all.

f x x x x x x x x x

x k i

ik

( ) ( )( )(

) .( )( )

1 1 1

1

3 5 3 9 15 5 15

2 1 2 1

00

Ferrer's graph

14=4+3+3+2+1+1=6+4+3+1

The number of partitions ofan integer n into m summandsis equal to the number ofpartitions where m is the largest summands.


Chapter 99.4 The Exponential Generating Function

ordinary generating functions: selections (order is irrelevant)

( ) ( , ) ( , )!

( )

1

1 00

0 0

x C n r x P n rxr

x C n C nP n P n

n r

r

n r

r

n

n is the ordinary generating function of ( , ), ( ,1),But an exponential generating function for ( , ), ( ,1),

Def. 9.2 For a sequence of real numbers,

( ) =

is called the exponential generating function for the givensequence.

Ex. 9.22 is the ordinary g. f. of 1,1,12!

and

an exponential g. f. of 1,1,1, .

a a a

f x a a x ax

ax

axi

exi

i

i

i

xi

i

0 1 2

0 1 2

2

3

3

0

0

2 3

13

, , ,

! ! !

!,

!,



Ex. 9.23 In how many ways can four of the letters in ENGINEbe arranged?For the letter E and N, we use ( + + because thereare 0, 1, or 2 to arrange. For G and I, we use (1 + ).Consequetly, the exponential generating function (of

arrangement) is ( ) = + + And the answer

is the coefficient of !

in ( ).

12

12

1

4

2

2 22

4

xx

x

f x xx

x

xf x

!)

!( ) .



Ex. 9.24 e xx x

e xx x

e e x x

e ex

x x

x

x

x x

x x

12 3

12 3

21

2 4

2 3 5

2 3

2 3

2 4

3 5

! !

! !

! !

! !

Ex. 9.25 A ship carries 48 flags, 12 each of the colors red, white,blue, and black. Twelve of these flags are placed on a verticalpole in order to communicate a signal to other ships.



Ex. 9.25 (continued)(a) How many of these signals use an even number of blue flags andodd number of black flags?

( ) ( )! ! ! ! !

! !

a

the coefficient of in ( ) yields 411.

f x x x x x x x x

e e e e e e e e e

xi

x f x

xx x x x

x x x x

i

i

12

12 4 3 5

2 214

14

1

14

412

2 2 2 4 3 5

2 2 2 2 4

1

12



Ex. 9.25 (continued)(b) How many of these signals have at least three white flags orno white flags at all?

( ) ( )! ! !

!( )

!

! !

!

b

the coefficient of in ( ) yields 10,754,218.

g x xx x x

e e xx

e xex e x

i

xxi

x xi

x g x

x x x xx i

ii

i

i

i

12

13 4

2 24

32

3

12

2 3 3 4

3 24 3

2 3

0

0

2

012


Chapter 9Ex. 9.26 Assign 11 new employees to 4 subdivisions. Eachsubdivision will get at least one new employees..

Each subdivision can be selected from 1 time to 8 times.

Hence the e.g.f. is ( ) +!

The

coefficient of !

is what we want. It is easier to work with

+!

The coefficient of !

is

411

f x xx x x

x

x x x x e e e e

e x

ii

x x x x

x

i

i

2 3 8 4

11

2 3 8 44 4 3 2

11

11 11 11 11

0

4

2 3 8

11

2 3 81 4 6

4 111

4 3 6 2 4 1 14

4

! !.

! !

.

( ) ( ) ( ) ( ) ( )

the number of onto functions


Chapter 99.5 The Summation Operator

The prefix sum sequence of a sequence is Let ( ) be the generating

function of Then ( )-

Therefore, ( )-

is the g. f. of the prefix sum.sequence.

00

0

a a aa a a a a a f x

a a a f xx

a a x a x

x x a a a x a a a xf x

x

, , ,, , , .

, , , .

( ) ( )

1 20 1 0 1 2

1 2 0 1 22

20 0 1 0 1 2

21

1

1


Chapter 99.5 The Summation Operator

Ex. 9.27 Find a formula for 0

The g. f. of 0 is ( + )

( - )Therefore, the g. f. of 0 0 0 is

( + )

( - ) The coefficient of is what we wnat.

( + )

( - )

2

2

2 2 2

1 2

1 2 1

11 1 2

1

11

11

40

41

42

2 2 2

2 23

2 2 2

4

42 4

2 2

nx x

x

x x

xx

x x

xx x x

x x x x

n

.

, , , .

, , ,

.

( )( )

( )

Ansn n

n n n

n n: ( ) ( )

( )( )

41

142

1

1 2 16

1 2


Chapter 9Summaries (m objects, n containers)

Objects Containers Some Number Are Are Containers ofDistinct Distinct May Be Empty Distributions Yes Yes Yes nm Yes Yes No n!S(m,n) Yes No Yes S(m,1)+S(m,2)+...+S(m,n) Yes No No S(m,n) No Yes Yes No Yes No No No Yes (1) p(m), for n=m No No No (2) p(m,1)+p(m,2)+...+p(m,n), n<m p(m,n)

n mm

1n m n

m n

( ) 1

p(m.n):number of partitions of m into exactly n summands


Chapter 9

Exercise: P390: 6 P399: 18,20 P403: 9,10 P408: 6

discrete math by r.s. chang, dept. csie, ndhu1 chapter 9

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