Chapter 2 Combinatorial Analysis 主講人 : 虞台文. Content Unordered Samples without Replacement Combinations Binomial Coefficients Some Useful Mathematic

Chapter 2Combinatorial Analysis

主講人 :虞台文

ContentUnordered Samples without Replacement Combinations

Binomial Coefficients Some Useful Mathematic ExpansionsUnordered Samples with ReplacementDerangement Calculus


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.

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.

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| |

m n

r r

1( ) | |

| |E EP

| |n

Em

k

m

m

r r

k

( )n

m

k kE

r

n

m

Pr

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

( )n

m

k kE

r

n

m

Pr

樂透和本例有何關係 ?

樂透和本例有何關係 ?

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.

nk

k

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 n

x

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 1 k

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

3

6 4 11 1

1z z

z

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

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

Documents

Chapter 2 Combinatorial Analysis 主講人 : 虞台文. Content Unordered Samples without Replacement Combinations Binomial Coefficients Some Useful Mathematic