Math Diskrit

7/31/2019 Math Diskrit

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CHAPTER 1

GENERATING FUNCTION


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Infinite series in the form

The series is called Power.

Taylor series function f (x) around x = 0 is:

~

0n

n

nxa

~

0 !.0

n

nn

n

xfxf

....!30'''!2.0''.0'0

32

x

f

x

fxff


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Five Formula with Taylor series

1rmula........Fo....................n!

xe

~

0n

nx

2rmula........Fo....................xx1

1 ~

0n

n

3ormula.........F....................0Zk;xn

1nk

x1

1 n~

0n

k

4ormula.........F..............................x1

x1x...xxx1

1nn32

5.ala.....Formu..............................ee21......

6!x

4!x

2!x1 xx

642

5.bla.....Formu..............................ee2

1......

7!

x

5!

x

3!

xx xx

753


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Formula for Combination

If then :

with:

1, xRt

~

0

1n

nt xn

tx

1,

!

1...21

0,1

n

n

ntttt

n

n

t

~

0 0

~

0

~

0

.n

nn

k

knk

n

n

n

n

n

n xbaxbxaxBxA


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Definition of generating function

Let (an) = (a1, a2, a3, ...) is a sequence of realnumbers.

Ordinary Generating Function (FPB) from an is:

Exponent Function Generator (FPE) from an is:

~

0n

n

nxaxP

~

0 !.

n

n

nn

xaxP


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Generating functions for Combinations

Suppose there are three different objects, a, b,

and c. How many ways to pick n objects with a

drawn object condition at most 1, b are drawn

at most 3, and c are being picked up at most 2.

This form is called the ordinary generating

function of the problem.

232 111 xxxxxxxP


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Examples problem of application PROBLEM

Determine the number of ways n letters taken from theword MATEMATIKA on the condition of each vowel andconsonant must be picked up at most 10 M drawn, and theconsonants K drawn at least 5 and at most 15. specify thenumber of ways n letters decision.

COMPLETIONNote that the word MATEMATIKA there are 6 differentletters are M, T, K, A, E, I and the letters are vowels A, E, I.Since each vowel should be drawn, each vowel is

associated with a factor (x + x2 + x3 + ...) in the generatingfunction.Consonant M can be chosen at most 10, then theconsonant M associated with a factor (1 + x + x2 + ... + x10).Consonants K can be chosen at least 5 and at most 15

associated with a factor (x5 + x6 + x7 + ... + x15)


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Generating function of the problem are:

1,,1550terambil10terambil

.........1...1332157652102

IEAKTM

xxxxxxxxxxxxxP

3102511

...1...11

1

1

1

xxxxxxx

xx

xxP

6

2118

111

x

xx

kn

k

xk

kxxx

0

3019816

2

30

0

19

0

8

0

5525

k

n

k

k

n

k

k

n

k

xk

kxk

kxk

k

nn

n

n

n

n

xn

nx

n

nx

n

nxP

~

30

~

19

~

8 30

25

19

142

8

3


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Determine the number of k combinations of n distinct objects if:a). Repetition is not allowedb). Repetition is allowed

COMPLETIONSuppose the object is O1, O2, O3, ...

a) Repetition not allowedBecause repetition is not allowed then each object is associatedwith a factor (1 + x) the generating function.Generating function of the above problems are:

Thus the number of combinations of k out of n objects withoutrepetition

xxxxP 1....11

~

0k

kx

kn

nk

k

n

0;


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b). Repetition is allowed.

Since repetition is allowed, each object is associated with afactor (1 + x + x2 + x3 + ...) in the generating functionGenerating function of the problem is

Thus the number of combinations of k out of n objects withrepetition is the coefficient of xk in P (x), namely: thecoefficient of xk in P (x), namely:

nxxxxP ...1 32 n

x

1

1

kn

k

xk

kn

0

1

k

kn 1


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Examples problem of application

PROBLEMDetermine the number of ways to put n identical ballsinto k different boxes on the condition that thereshould be no empty boxes.COMPLETION:

Since there are k different boxes and each box can notbe empty, then the function generator from the aboveproblems are:

kxxxxP 32)(

tkt

xt

tk

~

0

1

n

kn

x

kn

nxP

~ 1)(


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PROBLEMGiven equation.

Determine the number of solutions of the equation!

COMPLETION:

Zzyxdanzyxzyx ,,;2,50,0.10

...1...1 43254322 xxxxxxxxxxxP

n

n

n

n

xn

nx

n

n

~

8

~

2 8

6

2

kk

xk

kxx

~

0

8213

8~

0

2~

0

22

k

k

k

k

xk

kxk

k


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Generating Function for Permutation

Given three different letters a, b , and c.How many passwords of length n which can beformed from a, b , and c such that in everypassword: Point a appears at most 1, letter bappears at most 3 and Point c appears at most 2

completion:

This is the generating function of the manyproblems of passwords of length n is thecoefficient of:

!2!11

!3!2!11

!11

232 xxxxxxxP

xP

n

xn

dalam

!


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Determine the number of passwords of length n formed from

the word combinatorics such that each vowel must appear in

every password.

COMPLETIONGenerating function of the problem are:

3

326

32

.. .!3!2!1

.. .!3!2!1

1

xxxxxxxP

36 1 xx ee

~

0

~

0

~

0

~

0 !

6

!

73

!

83

!

9

n

nn

n

nn

n

nn

n

nn

n

x

n

x

n

x

n

x


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Let S be the set of n rows of binary numbers. If a line iswritten at random what is the probability of thesequence contains "0" as many as odd and "1" as muchas the even!

COMPLETIONSuppose that:S = The set of n rows of binary numbers.A = set of all binary n rows contain the numbers "0" andodd numbers "1" as much as the even

Asked: P (A)Exponential generating function to find n (A):

...

!4!21...

!3!2!1

4253 xxxxxxP

xxxx eeee 2121


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Exponential generating function to find n (S):

So the chances of the sequence contains "0" as much

as even-odd and 1 is:

~

0

~

0 !.2

!.2

4

1

n

nn

n

nn

n

x

n

x

ganjiln

genapnAn

n ;2

;0

1

2

432

...!4!3!2!1

1

xxxxxP

~

0 !.2

n

nn

n

x 02 nSn n

snAn

AP

ganjiln

genapn

;2

1

;0


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Examples problem of application Determine the number of ways to put n balls into k boxes

such that no box is empty if:Different balls in different boxes

Different balls in a box identical

Since n different balls in k different boxes and no boxes are

empty then the generating function exponent is:

.. .

!3!2!1...... .

!3!2!1...

!3!2!1

321321321 xxxxxxxxxxP

k

xxxxP

.. .

!3!2

32

xtkxtktk

t

eet

k

inibagianekspansi;1

0

~

00 !.1

n

nnt

k

t n

xtk

t

k

k

t

nt

n tkt

ka

0

1


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If the box is identical to the number of way in

question is the answer to number a) above,

divided by K!.

So many ways to put an object with no empty

box is:

k

t

nttk

t

k

kknS

0

1!

1,


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Chapter 2

Recursive Relation


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Recursive Relation Linear Recursive Relations with Constant Coefficients

The general form of recursive part, of a linear recursive relation of

degree k is as follows:

an + h1 (n) an-1 + h2 (n) an-2 + h3 (n) an-3 + ... + Hk (n) an-k = f(n)

with hi (n) and f (n) is a function in and hk(n) 0

If f (n) = 0 then the relation rekursifnya called homogeneous, otherwisecalled nonhomogeneous.

Homogeneous Linear Recursive Relations with Constant Coefficients

The general form of linear homogeneous recursive relation with

constant coefficients are as follows:

c1an + an-1 + ... Ckan +-k = 0; ck 0 (1.2.1)

with k initial conditions, and for 1 i k, ci = constant


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Superposition Theorem

Theorem 1.2.1 (Principle of Superposition):

If g1 (n) the solution of an + c1an-1+ c2an-2+ c3an-3 + ... + ckan-k=f1(n)

and g2(n) solution of

an + c1an-1+ c2an-2+ c3an-3 + ... + ckan-k=f2(n)

then the linear combination

k1 g1(n) + k2 g2(n)

solution of an + c1an-1+ c2an-2+ c3an-3 + ... + ckan-k= k1f1(n) + k2 f2(n).


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Resolving Relationship With Recursive Linear

Homogeneous Constant Coefficient

The general form of linear homogeneous recursive relation with constantcoefficients are:an + c1an-1 + c2an-2 + ... + ckan-k= 0 ; n k... the recursivea0 = p0, a1 = p1 , ... ak-1 = pk-1 ... terms / conditions of the initial

Completion steps:

Let an = xn, x 0

2) Section rekursifnya be:xn + c1xn-1 + c2x

n-2 + ... + ckxn-k= 0

Smallest rank is n - k

3) For both sides of the equation by xn-k, is obtained:xk+ c1xk-1 + c2xk-2 + ... + ck= 0

The equation above is an equation called the "EquationCharacteristics "of the recursive relation above. In the general recursive

relationhas k roots. Suppose the roots of the characteristic equation isx1, x2, ... , Xk.


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CASE I: CHARACTERISTICS OF DIFFERENT ALL EQUAL ROOTS

In this case x1n, x2

n, ... , xkn is the solution of the above

recursive relation, so as a result of the superposition theorem

is obtained:k1x1

n + k2x2n + ... + k

kxkn solution of the recursive relation.

So, the solution of the recursive relation is:

an

= k1x1n+ k2x2

n+ ... + kkxkn

Case II: ROOTS THERE COPIES characteristic equation

If the characteristic equation xk+ c1xk-1+ c2x

k-2+ ... + ck

= 0 of

the recursive relation an + c1an-1 + ... + ckan-k = 0 ; ck 0 has

a root, say x1 (double root m, m k) then the generalsolution of recursive relationships involving x1 has the form:

c1x1n+ c2nx1

n+ c3n2x1

n+ c4n3x1

n+...+ cmnm-1x1

n


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Examples problem of application Find the formula of the n numbers in the Fibonacci sequence.

Recursive relation for Fn are:Fn = Fn-1 + Fn-2, n 2

F0 = F1 = 1

Step-by-step solution:

Let Fn = xn; x 0 then the recursive form of Fn - Fn-1 - Fn-2 = 0

becomes:

xn - xn-1 - xn-2 = 0

for both sides of this last characteristic equation xn-2with the

characteristic equation is obtained as follows:

x2 - x - 1 = 0


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Look for a formula to satisfy the following relationan-= 3an-1 + 6an-2 - 28an-3+24an-4with a0 = 1 a1 = 2 a2 = 3 and a3 = 4

completion:

Let an = xn

; x 0 then the form of recursivexn 3xn-1 6xn-2 + 28xn-3 24xn-4 = 0

For both sides of this last characteristic equation xn-4 obtainedthe characteristic equation as follows:x4 3x3 6x2 + 28x 24 = 0

Equivalent to (x - 2)3 (x + 3) = 0

an =c12n+ c2n2

n+ c3n22n+ c4(-3)

n


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Resolving Recursive Relation With

Generating Function

In previous chapters we have been talking

about generating functions and applications.

In this section we will see that the generating

function can also be used to find a recursivesolution easily.

Examples of:

Use the ordinary generating function to solvethe following recursive relation:

a0 = 1, a1 = 3; an = 2an-1 + 4n-1, n 2


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COMPLETION Suppose P (x) is the ordinary generating function of

the sequence (an), then by definition:

Since for n 2, an = an-1 + 2n-1+4n-1, if both sides of

this equation xn multiplied then summed for n = 2 ton = , obtained

So that equation (1.3.1) becomes

P(x) - 1 - 3x=2xP(x) - 2x +


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DERANGEMENT

Suppose there are n elements align in one row and labeled1,2,3, ... .., n. Then the n elements in the row

dipermutasikanThe same such that there is no one occupying elemenpun

its original position. A permutation is calledDerangement.Recursive Relation for Dn is:

The formula for Dn is


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Recursive Relation System

ISSUES.....

Suppose the stated amount of n-digit binary sequence thatincludes:"0" as much as an even number and "1" as much as an even

number;

bn stating the number of binary n-digit sequence thatcontains "0" as much as even and "1" as many as odd;

cn is the number of binary n-digit sequence that contains"0" as many as odd and "1" as much as fulfilled, and

dn is the number of binary n-digit sequence that contains"0" as many as odd and "1" as many as odd.


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Problems solutions

Since each n-digit binary sequence whichcontains "0" as much as even and "1" even asmuch as can be obtained from: a binarysequence (n-1)-numbers that contain "0" as

much as even and "1" as many as odd byadding / insert a digit "1", or a binarysequence (n-1)-numbers that contain "0" asmany as odd and "1" as much even with the

add / insert a digit "0",then obtained the following relationship:


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Similarly, any n-digit binary sequence which contains

"0" as much as even and "1" as many as odd can beobtained from:

a binary sequence (n-1)-numbers that contain "0" as

much as even and "1" as much even by inserting a

digit "1"; ora binary sequence (n-1)-numbers that contain "0" as

many as odd and "1" as many as odd by inserting a

digit "0". Thus obtained the following relationship:


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With a similar argument can be shown that for

and, successively apply the followingrelationship:

Furthermore, the

If a0 = 1, b0 = c0 = d0 = 0, then:


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Next, we use generating functions to solve the

recursive system. Suppose that A (x), B (x), C(x) and D (x), respectively, ordinary generating

function of an, bn, cn, and dn, is obtained:


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Chapter 3

Principle of inclusion-exclusion


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Principle of inclusion-exclusion

Theorem 3. 2. 1: (Principle of inclusion-exclusion)

If N is the number of objects in a set S and a1,. . . , ar

properties - properties that may be possessed by anobject in S, then the number of objects in S that do

not have the properties of a1, a2,. . . , Ar is


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Many Objects That Have Exactly m

Properties

Theorem 3.3.1

Let a1, a2, ..., ar are traits that may be

possessed by an object in the set S, then a lot

of objects S which has exactly m r propertiesare


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Many objects that have properties of

total Even or Odd

Theorem 3.4.1:

If in the set S there is r nature, then the number of

objects that S has an even number of properties are:

S and the number of objects that have properties of

an odd number is


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PROBLEM1. There is some integer from 1 to 1000 thatNot divisible by 3 or 5?Not divisible by 3, 5, or 7?

COMPLETIONLet S = {1,2,3, ..., 1000}.a1: the nature divisible by 3;a2: the nature divisible by 5;

a3: nature is divisible by 7.asked are:

(a). (b).


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Completion

It is clear that. Next we get,

N (a1) = number of members of S is divisible by 3 = 333

N (a2) = number of members of S is divisible by 5 = 200

N (a3) = number of members of S is divisible by 7 = 142N (a1a2) = number of members of S is divisible by 3 and

5 = 66

N (a2a3) = 28

N (a1a3) = 47

N (a1a2a3) = 9


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Thus, the inclusion-exclusion principle, beobtained

= 1000 - 333 - 200 - 66 = 533;

= 1000 - 333 - 200 - 142 + 66 + 47 + 28 9

= 457.


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A total of n different balls are placed into k differentboxes. What is the probability that there is an empty

box?

COMPLETION

Suppose S is the set of all events (distribution) is

possible. Ei be the event that the empty box to i and

ai is the nature of that event Ei appears. In this case i {1,2, ..., k}


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Use the principle of inclusion-exclusion to determinebanyaknsolusi round from the following equation:

x1 + x2 + x3 = 20, 0 xi 5, i {1,2,3}

completion:

Let S be the set of all round solution of the equation x1

+ x2

+

x3= 20, xi 0 i {1,2,3}. Then it can be shown that

According to the principle of inclusion-exclusion, is obtained:


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A total of n pairs of husband and wife attending adance party. Classes held simultaneously and aman dancing with a woman.

a) What is the probability there is exactly onehusband and wife pairs dancing togetherin the dance?

b) What opportunities there are exactly threehusband and wife pairs dancing togetherin the dance?


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Completion

Let S be the set of all possible dance partner, and indicate the

nature ai where i is paired with her husband to his wife.

1 i n. Since there are n pairs of husband and wife, then

N = S = n!.

Furthermore, we obtainOpportunities lie just one pair of husband-wife dance

together is:

opportunities there are exactly three pairs of husband-wife

dance together is:

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