chapter 2 combinatorial analysis 主講人 : 虞台文. content basic procedure for probability...

Chapter 2Combinatorial Analysis

主講人 :虞台文

Content Basic Procedure for Probability Calculation Counting

– Ordered Samples with Replacement– Ordered Samples without Replacement Permutations– Unordered Samples without Replacement Combinations

Binomial Coefficients Some Useful Mathematic Expansions Unordered Samples with Replacement Derangement Calculus


Basic Procedure for

Probability Calculation

Basic Procedure for Probability Calculation

1. Identify the sample space .2. Assign probabilities to certain events

in A, e.g., sample point event P().3. Identify the events of interest.4. Compute the desired probabilities.


Counting

Goal of Counting

Counting the number of elements in a particular set, e.g., a sample space, an event, etc.

? E

?E

Cases

Ordered Samples w/ Replacement

Ordered Samples w/o Replacement

– Permutations

Unordered Samples w/o Replacement

– Combinations

Unordered Samples w/ Replacement


Ordered Samples

with Replacement

Ordered Samples

eat

ate

tea

elements in samples appearing in different orders are considered different.


meet

teem

mete

1. Elements in samples appearing in different orders are considered different.

2. In each sample, elements are allowed repeatedly selected.


Drawing k objects, their order is noted, among n distinct objects with replacement.

The number of possible outcomes is

ndistinctobjects

k

kn

Example 1

How many possible 16-bit binary words we may have?

2distinctobjects

16 1 16 {0,1}, 1, ,16ia a a i

162

Example 2

Randomly Choosing k digits from decimal number, Find the probability that the number is a valid octal number.

Randomly Choosing k digits from decimal number, Find the probability that the number is a valid octal number.

1 {0, ,9}, 1, ,k ia a a i k 10k

For any , P()=1/10k.

1 {0, ,7}, 1, ,k iE b b b i k 8kE

1( ) 8

10k

kP E 0.8k


Ordered Samples

without Replacement

Permutations

Permutations

清心

也可

以

可以清心也以清心也可清心也可以心也可以清也可以清心

可以清心也以清心也可清心也可以心也可以清也可以清心

Ordered Samples w/o Replacement Permutations

Drawing k objects, their order is noted, among n distinct objects without replacement.

The number of possible outcomes is

ndistinctobjects

k

( 1)( 2) ( 1)n n n n k nkP

!

( )!

n

n k

Example 3

Five letters are to be selected without replacement from the alphabet (size 26) to form a word (possibly nonsense). Find the probabilities of the following events?1. Begin with an s.2. Contains no vowel.3. Begins and ends with a consonant.4. Contains only vowels.


Example 3



Define

E1: word begins with an s.E2: word contains no vowel.E3: word begins and ends with a consonant.E4: word contains only vowels.

P(E1)=? P(E2)=? P(E3)=? P(E4)=?

Example 3Define

E1: word begins with an s.E2: word contains no vowel.E3: word begins and ends with a consonant.E4: word contains only vowels.

P(E1)=? P(E2)=? P(E3)=? P(E4)=?

1 2 3 4 5

{ , , }

,i

i j

a a za a a a a

a a i j

265P

For any , P()=1/||.

26 25 24 23 22

1 1 25 24 23 22E

2 21 20 19 18 17E

3 21 24 23 22 20E

4 5 4 3 2 1E

1 1

1( )

| |P E E

2 2

1( )

| |P E E

3 3

1( )

| |P E E

4 4

1( )

| |P E E

0.0385

0.3093

0.6462

21.52 10


Unordered Samples

without Replacement

Combinations

Combinations

n distinctobjects

Choose k objectsChoose k objects

How many choices?How many choices?

Combinations

Drawing k objects, their order is unnoted, among n distinct objects w/o replacement, the number of possible outcomes is

nkC

n

k

( 1) ( 1)

!

n n n k

k

!

( )! !

n

n k k

This notation is preferred

n distinctobjects

Choose k objectsChoose k objects

How many choices?

How many choices?

More onn

k

( 1) ( 1)0

!0 0

n n n kn k

kk

k

Examples

( 1) ( 1)0

!0 0

n n n kn k

kk

k

2.5

3

2.5 1.5 0.5

3!

1

3

1 2 3

3!

1

0.3125

Example 4

The mathematics department consists of 25 full professors, and 15 associate professors, and 35 assistant professors. A committee of 6 is selected at random from the faculty of the department. Find the probability that all the members of the committee are assistant professors.

The mathematics department consists of 25 full professors, and 15 associate professors, and 35 assistant professors. A committee of 6 is selected at random from the faculty of the department. Find the probability that all the members of the committee are assistant professors.

x

Denoting the all-assistant event as E,

75

6

35

6E

35 751( )

6 6P E E

Example 5

A poker hand has five cards drawn from an ordinary deck of 52 cards. Find the probability that the poker hand has exactly 2 kings.

A poker hand has five cards drawn from an ordinary deck of 52 cards. Find the probability that the poker hand has exactly 2 kings.

x

Denoting the 2-king event as E,

52

5

4 48

2 3E

4 48

2 31( )

52

5

P E E

Example 6

Two boxes both have r balls numbered 1, 2, … , r. Two random samples of size m and n are drawn without replacement from the 1st and 2nd boxes, respectively. Find the probability that these two samples have exactly k balls with common numbers.

Two boxes both have r balls numbered 1, 2, … , r. Two random samples of size m and n are drawn without replacement from the 1st and 2nd boxes, respectively. Find the probability that these two samples have exactly k balls with common numbers.

11 22 33 rr

11 22 33 rr

m

n P(“k matches”) = ?

E

||=?

|E|=?

1( ) | |

| |E EP

Example 6

11 22 33 rr

11 22 33 rr

m

n| |

m n

r r

1( ) | |

| |E EP

| |n

Em

k

m

m

r r

k

# possible outcomes from the 1st box.

# possible k-matches.

# possible outcomes from the 2nd box for each k-match.

Example 6

11 22 33 rr

11 22 33 rr

m

n

1( ) | |

| |E EP

( )n

m

k kE

r

n

m

Pr

本式觀念上係由第一口箱子

出發所推得本式觀念上係

由第一口箱子出發所推得

Example 6

11 22 33 rr

11 22 33 rr

m

n

1( ) | |

| |E EP

( )nk k

m

n

r

Pr

m

E

觀念上，若改由第二口箱子

出發結果將如何 ?

觀念上，若改由第二口箱子

出發結果將如何 ?

Example 6

11 22 33 rr

11 22 33 rr

m

n

1( ) | |

| |E EP

( )nk k

m

n

r

Pr

m

E

k km

r

m

r

n n

Exercise11 22 33 rr

11 22 33 rr

m

n

1( ) | |

| |E EP

( )nk k

m

n

r

Pr

m

E

k km

r

m

r

n n

r r

r

k k

m m n n

n m

m

k k

n

r

證明

r r

r

k k

m m n n

n m

m

k k

n

r

證明


Binomial Coefficients


0

1

2 2 2

3 3 2 2 3

4 4 3 2 2 3 4

1

2

3 3

4 6 4

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


0

1

2 2 2

3 3 2 2 3

4 4 3 2 2 3 4

1

2

3 3

4 6 4

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

?n

x y

Binomial Coefficients ?

nx y

nx y x y x y x y x y

1 2 2? ? ?n kn nkn nx x y x y x y y

n terms

Binomial Coefficients ?

nx y


1 2 2? ? ?n kn nkn nx x y x y x y y

n boxes

1

n 2

n

n

k

0

n

n

n



0

nn k k

k

yn

kx

0

nk n k

k

n

kx y

Facts: 0 for 1. n

k nk

0 for 02.n

kk

0

n k k

k kx y

n

0

k n k

k

n

kx y

n k k

k

xk

yn

k n k

k

n

kx y

Properties of Binomial Coefficients


0 0

1 1n n

k n k

k k

n n

k k

0

nn k n k

k

x y x yn

k

1 1n

2n


0 0

1 1 1n n

k k n k

k k

n n

k k

0

nn k n k

k

x y x yn

k

1 1n

0


0

nn k n k

k

x y x yn

k

Exercise


n 不同物件任取 k 個

第一類取法 :

第二類取法 :

1n

k

1

1

n

k


1 1

1

n n n

k k k

1

0

1

1

2

0

2

1

2

2

0

0

3

0

3

1

3

2

3

3

4

0

4

1

4

2

4

3

4

4

Pascal Triangular


1

0

1

1

2

0

2

1

2

2

0

0

3

0

3

1

3

2

3

3

4

0

4

1

4

2

4

3

4

4

Pascal Triangular

1

1

1

1

1

1

1

1

1

2

3 3

4 6 4

1 1

1

n n n

k k k


　吸星大法

1

1

n nk n

k k

Example 7-1

0

nk

k

nx

k

0

1n

k n k

k

nx

k

1

nx

Example 7-2

0

n

k

nk

k

0

k n

k

nk

k

1

k n

k

nk

k

1

1

1

k n

k

nn

k

1

1

1

k n

k

nn

k

kx+11

1 1

1

1 1

x n

x

nn

x

1

0

1x n

x

nn

x

1

0

1k n

k

nn

k

1

0

1n

k

nn

k

12nn

Fact:0

2n

n

k

n

k

?

Example 7-2

0

n

k

nk

k

1

k n

k

nk

k

1

1

1

k n

k

nn

k

kk+1

1

0

1n

k

nn

k

12nn

簡化版

Example 7-3

0

( 1)n

k

nk k

k

2

( 1)k n

k

nk k

k

kk+2

2( 1)2nn n

簡化版

2

1( 1)

1

k n

k

nn k

k

2

2( 1)

2

k n

k

nn n

k

2

0

2( 1)

n

k

nn n

k

?

Negative Binomial Coefficients

11

k

k k

n kn


11

k

k k

n kn

How to memorize?

1 1

kn k

k k

n

k

k k (n) 1


11

k

k k

n kn

這公式真的對嗎 ?

1 1

k

k

n

k

k k (n+k1) 1 1

k 1k

1

k

n


( 1) ( 1)

!

n n n k

k

k

n

( 1) ( 1)1

!k kn n n

k

( 1) ( 1)1

!k n

k

nk n

11

k kn

k


Some Useful Mathematic Expansions


1 1 z1

1 zz

z

2z z2z2 3z z

2z

3z


1 1

1 1 ( )z z

2 431 ( ) )(( )) (z z zz

2 3 41 z zz z


21

1 z

211

1z z

z

2 3 41 ? ? ? ?z z z z

2 21 1z z z z


21

1 z

211

1z z

z

2 3 41 2 3 4 5z z z z

2 21 1z z z z

31

1 z

2 3 41 ? ? ? ?z z z z

2 2 21 1 1z z z z z z

1

1

n

z

2 3 41 ? ? ? ?z z z z


211

1z z

z

1

1

n

z

1 ( )n

z

0

( ) 1k n k

k

nz

k

( ) 1n

z

0

( 1)k k

k

nz

k

0

1( 1) ( 1)k k k

k

n kz

k

0

1 k

k

n kz

k


2 2 1 2 31 1k k k kz z z z z z z z

2 1 21 1kz z z z z 11

1 1

kz

z z

11

1

kz

z

z值沒有任何限制


Unordered Samples with Replacement

Discussion

投返非投返

有序

無序

knn

kP

n

k

?


n不同物件任取 k個可重複選取

n不同物件，每一中物件均無窮多個從其中任取 k個


0 1 2 3z z z z 0 1 2 3z z z z 0 1 2 3z z z z

此多項式乘開後zk之係數有何意義 ?


0 1 2 3z z z z 0 1 2 3z z z z 0 1 2 3z z z z

2 3 2 3 2 31 1 1z z z z z z z z z

2 31n

z z z

1

1

n

z

0

1 k

k

n k

kz


投返非投返

有序

無序

knn

kP

n

k

1n k

k

Example 8

Suppose there are 3 boxes which can supply infinite red balls, green balls, and blue balls, respectively. How many possible outcomes if ten balls are chosen from them?

n = 3k = 10

3 10 1 12

10 10

Example 9

There are 3 boxes, the 1st box contains 5 red balls, the 2nd box contains 3 green balls, and the 3rd box contains infinite many blue balls. How many possible outcomes if k balls are chosen from them.

k=1有幾種取法k=2有幾種取法k=3有幾種取法k=4有幾種取法

觀察 :

Example 9

2 3 4 51 z z z z z 2 31 z z z 2 31 z z z

此多項式乘開後zk之係數卽為解

Example 9

2 3 4 51 z z z z z 2 31 z z z 2 31 z z z

6 41 1 1

1 1 1

z z

z z z

36 4 1

1 11

z zz

3

4 6 10 11

1z z z

z

4 6 10

0

3 11 j

j

jz z z z

j

4 6 10

0

21 j

j

jz z z z

j


Example 9


4 6 10

0

21 j

j

jz z z z

j

0 0 0 0

4 6 102 2 2 2j j j j

j j j j

z zj j j j

z z zj j j j

z z

0 0 0

4 10

0

62 2 2 2j j j j

j j j j

j j j jz z z z

j j j j

Coef(zk)=?

Example 9

4 6 1

0

0

0 0 0

2 2 2 2j j j j

j j j j

j j j jz z z z

j j j j

Coef(zk)=?

從 z0 開始從 z4 開始從 z6 開始從 z10 開始

0 4 6 10

2 2 4 8

4 6 10k k k k

k k k k

k k k kz z z z

k k k k

jk4 jk6 jk10jk

Example 9

0 4 6 10

2 2 4 8

4 6 10k k k k

k k k k

k k k kz z z z

k k k k

20 4

2 24 6

4( )

2 2 46 10

4 6

2 2 4 810

4 6 10

k

kk

k

k kk

k kCoef z

k k kk

k k k

k k k kk

k k k k

Example 9

20 4

2 24 6

4( )

2 2 46 10

4 6

2 2 4 810

4 6 10

k

kk

k

k kk

k kCoef z

k k kk

k k k

k k k kk

k k k k

2 2 4 8( )

4 6 10k k k k k

Coef zk k k k


Derangement

Derangement

最後 ! ! !每一個人都拿到別人的帽子

錯排

Example 10

nkE 0

nE

n人中正好 k人拿對自己的帽子n人中正好 k人拿對自己的帽子 n人中無人拿

對自己的帽子n人中無人拿對自己的帽子

). ?3 ( nkEP

0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

Example 10

nkE 0

nE

). ?3 ( nkEP

0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

? ?nkE 0 ?nE

Example 10nkE

0nE

). ?3 ( nkEP

0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

n人中正好 k人拿對自己的帽子 n人中無人拿對自己的帽子

!n

20( ) ?EP 3

0( ) ?EP

1 2

2 1

1 2

2 3

3

1

3 1 2

1/2! 2/3!

Example 10nkE

0nE

). ?3 ( nkEP

0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

n人中正好 k人拿對自己的帽子 n人中無人拿對自己的帽子

!n

令 Ai表第 i個人拿了自己帽子

0 0( ) 1 ( )n nP E P E 1 21 ( )nP A A A

Example 10 ). ?3 ( nkEP

0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

Ai表第 i個人拿了自己帽子

0 1 2( ) 1 ( )nnP E P A A A


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E


0 1 2( ) 1 ( )nnP E P A A A

1 2 n

1

( 1)!( )

!i

nP A

n

. . .

. . .


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E


0 1 2( ) 1 ( )nnP E P A A A

1 2 n

1 2

( 1)!( )

!i

nP A

n

. . .

. . .

( 2)!( ) ,

!i j

nP A A i j

n


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E


0 1 2( ) 1 ( )nnP E P A A A

( 1)!( )

!i

nP A

n

( 2)!( ) ,

!i j

nP A A i j

n

( 3)!( ) ,

!i j k

nP A A A i j k

n

1

n

計項1

n

計項2

n

計項2

n

計項

3

n

計項3

n

計項

( )!

!k

n n kS

k n

! ( )!

!( )! !

n n k

k n k n

1

!k


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E


0 1 2( ) 1 ( )nnP E P A A A

( )!

!k

n n kS

k n

! ( )!

!( )! !

n n k

k n k n

1

!k 1

0

1 1 1( ) 1 1 1

2! 3! !nnP E

n

1

0

1 1 1( ) 1 1 1

2! 3! !nnP E

n

11 1 11 1

2! 3! !n

n


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E


0 1 2( ) 1 ( )nnP E P A A A

1

0

1 1 1( ) 1 1 1

2! 3! !nnP E

n

1

0

1 1 1( ) 1 1 1

2! 3! !nnP E

n

20

1 1( ) 1 1

2! 2P E

30

1 1 1( ) 1 1

2! 3! 3P E

40

1 1 1 3( ) 1 1

2! 3! 4! 8P E


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E


0 1 2( ) 1 ( )nnP E P A A A

1

0

1 1 1( ) 1 1 1

2! 3! !nnP E

n

11 1 1 11 1

1! 2! 3! !n

n

0

1 1 1 1( ) 1 1

1! 2! 3! !nnP E

n


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

Ai表第個人拿了自己帽子

0 1 2( ) 1 ( )nnP E P A A A

1

0

1 1 1( ) 1 1 1

2! 3! !nnP E

n

11 1 1 11 1

1! 2! 3! !n

n

10im ( )l

n

nEP e

10im ( )l

n

nEP e

0

1 1 1 1( ) 1 1

1! 2! 3! !nnP E

n


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

0

0( )!

n

nE

P En

( )!

nkn

k

EP E

n

0 0! ( )n nE n P E

?nkE

0

1 1 1 1( ) 1 1

1! 2! 3! !nnP E

n


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

0

0( )!

n

nE

P En

( )!

nkn

k

EP E

n

0 0! ( )n nE n P E

?nkE

. . .. . .

. . .. . .

k matches n k mismatches

n

k

0n kE 0

!

n kEn

k n

0

1( ) ( )

!n n kkP E P E

k

0( )!

! ( )!

n kEn n k

k n n k


0 ). ?1 ( nEP

0 ) ?2. lim (n

nP E

52( ) ?P E 3

0

1

2!P E

0

1( ) ( )

!n n kkP E P E

k

1 1

2! 3

Remark

0

( ?)n

nk

k

P E

1


Calculus

Some Important Derivatives

Derivatives for multiplications —

Derivatives for divisions —

Chain rule —

duvu v uv

dx

2

d v uv u v

dx u u

dy dy du

dx du dx ( )

( )

y f u

u g x

L’Hopital rule

Suppose as we hav( )

, .(

0e or

0)

f xx c

g x

( ) ( )lim lim

( ) ( )x c x c

f x f x

g x g x

Examples

Integration by Part

duv udv vdu

duv udv vdu uv udv vdu

Integration by Part

uv udv vdu udv uv vdu

b bb

aa audv uv vdu

The Gamma Function

1

0( ) , 0xx e dx

Example 12

1

0( ) , 0xx e dx

Example 12

1

0( ) , 0xx e dx

0(1) xe dx

0

xe

0 ( 1)

1

Example 12

1

0( ) , 0xx e dx

0

Example 12

1

0( ) , 0xx e dx

1

0( ) xx e dx

1

0

xx de

1 1

0 0

x xx e e dx 1 2

0 0( 1)x xx e x e dx

( 1)

( 1) ( 1)

Example 12

1

0( ) , 0xx e dx

( ) ( 1) ( 1)

( 1)( 2) ( 2)

( 1)( 2)( 3) (1)

( 1)!

Example 12

1

0( ) , 0xx e dx

2 / 2 ?xe dx

2 / 2 ?xe dx

Example 12

1

0( ) , 0xx e dx

2 / 2 ?xe dx

2 / 2 ?xe dx

2 2 2/ 2 / 2 / 2x x ye dx e dx e dy

2I

2 2

2 2

x y

I e dxdy

22 / 2

0 0

re rdrd

22

2 / 2

0 0 2r r

e d d

2

0d

2

2

Example 12

1

0( ) , 0xx e dx

2 / 2 2xe dx

Example 12

1/ 2

0

1

2xx e dx

2 / 2 2xe dx

1

0( ) , 0xx e dx

2et 2L /x y1/ 2

0

x x

xx e dx

222

2

2

1/ 22 2/ 2

0 2 2

y

y

yy ye d

2 / 2

0

2y y

ye ydy

y

2 / 2

02 ye dy

2 / 22

2ye dy

Example 12

1

0( ) , 0xx e dx

Example 12

1

0( ) , 0xx e dx

?3

2

?5

2

?7

2

1 1

2 2 2

3 3

2 2

3

4

5 5

2 2

15

8

chapter 2 combinatorial analysis 主講人 : 虞台文. content basic procedure for probability...

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