Chapter 2
Combinatorial 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
Chapter 2
Combinatorial Analysis
Basic Procedure for
Probability Calculation
Basic Procedure for Probability Calculation
1.
2.
3.
4.
Identify the sample space .
Assign probabilities to certain events in A,
e.g., sample point event P().
Identify the events of interest.
Compute the desired probabilities.
Chapter 2
Combinatorial Analysis
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
–

Unordered Samples w/o Replacement
–

Permutations
Combinations
Unordered Samples w/ Replacement
Chapter 2
Combinatorial Analysis
Ordered Samples
with Replacement
Ordered Samples
eat
ate
tea
elements in samples appearing
in different orders are
considered different.
Ordered Samples w/ Replacement
meet
teem
mete
1. Elements in samples appearing
in different orders are
considered different.
2. In each sample, elements are
allowed repeatedly selected.
Ordered Samples w/ Replacement
n
distinct
objects



k
Drawing k objects, their order is noted,
among n distinct objects with replacement.
The number of possible outcomes is
n
k
Example 1
How many possible 16-bit binary words we may have?
2
distinct
objects
  a1

16
a16 ai  {0,1}, i  1,
 2
16
,16
Example 2
Randomly Choosing k digits from decimal number, Find
the probability that the number is a valid octal number.
  a1
ak ai  {0,
,9}, i  1,
, k
   10k
For any , P()=1/10k.
E  b1
bk bi  {0,
, 7}, i  1,
1
 P( E )  8  k  0.8k
10
k
, k
 E  8k
Chapter 2
Combinatorial Analysis
Ordered Samples
without Replacement 
Permutations
Permutations
清
以
心
可
也
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
n
distinct
objects



k
Drawing k objects, their order is noted, among n
distinct objects without replacement.
The number of possible outcomes is
P  n(n  1)(n  2)
n
k
n!
(n  k  1) 
(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.
E1: word
Define E2: word
E3: word
E4: word
Example 3
P(E1)=?
begins with an s.
contains no vowel.
begins and ends with a consonant.
contains only vowels.
P(E2)=?
P(E3)=?
P(E4)=?
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.
E1: word
Define E2: word
E3: word
E4: word
Example 3
begins with an s.
contains no vowel.
begins and ends with a consonant.
contains only vowels.
P(E1)=?
P(E2)=?
P(E3)=?
P(E4)=?
26



P
 26  25  24  23 22

ai  {a, , z}
5
  a1a2 a3a4 a5

a

a
,
i

j
For any , P()=1/||.
i
j


E1  1 25  24  23 22
P( E1 ) 
E2  21 20 19 18 17
P( E2 ) 
E3  21 24  23  22  20
P( E3 ) 
E4  5  4  3  2 1
P( E4 ) 
1
E1 
||
1
E2 
||
1
E3 
||
1
E4 
||
 0.0385
 0.3093
 0.6462
 1.52 102
Chapter 2
Combinatorial Analysis
Unordered Samples
without Replacement 
Combinations
Combinations
n distinct
objects
Choose k objects
n distinct
objects
Combinations

yycchhooiciceess??
n
a
m
n
a
w
m
o
HHow
Choose
Choosekkobjects
objects
Drawing k objects, their order is unnoted, among
n distinct objects w/o replacement, the number
of possible outcomes is
 n  n(n  1) (n  k  1)
n!
C   

k!
(n  k )!k !
k 
n
k
This notation is preferred
n


More on  k 
 n(n  1) (n  k  1)
n 
k!
 
 k  0

k 0
k 0
Examples
 n(n  1) (n  k  1)
n
  
k!
 
k
  0

 1 1  2    3
 1
  
3!
3
 2.5  2.5  1.5  0.5
 0.3125
  
3!
 3 
k 0
k 0
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.



75
x   
6
 35 
Denoting the all-assistant event as E, E   
6
1  35   75 
P( E )  E 
   
 6 6
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.



52
x   
5
 4  48 
Denoting the 2-king event as E, E    
 2  3 
 4  48 
  
1  2  3 
P( E )  E 


 52 
 
5
Example 6
1 2 3
1 2 3
||=?
|E|=?
r
r
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.
m
n
P(“k matches”) = ?
E
1
P( E ) | E | 
||
1
P( E ) | E | 
||
Example 6
1 2 3
r
m
1 2 3
r
n
 r  r 
|  |   
 m  n 
 r  m  r  m 
| E |   

 m  k  n  k 
# possible outcomes from the 1st box.
# possible k-matches.
# possible outcomes from the 2nd box
for each k-match.
1
P( E ) | E | 
||
Example 6
1 2 3
r
m
1 2 3
r
n
 r  r 
|  |   
 m  n 
 r  m  r  m 
| E |   

 m  k  n  k 
 m  r  m 
 

k
n

k

P( E )   
r
 
n
1
P( E ) | E | 
||
Example 6
1 2 3
r
m
1 2 3
r
n
 r  r 
|  |   
 m  n 
 r  m  r  m 
| E |   

 m  k  n  k 
 m  r  m 
 

k
n

k

P( E )   
r
 
n
1
P( E ) | E | 
||
Example 6
1 2 3
r
m
1 2 3
r
n
 m  r  m 
 

k
n

k

P( E )   
r
 
n
1
P( E ) | E | 
||
Example 6
1 2 3
r
m
1 2 3
r
n
 m  r  m 
 

k
n

k

P( E )   
r
 
n
1
P( E ) | E | 
||
Example 6
1 2 3
r
m
1 2 3
r
n
 m  r  m   n  r  n 
 
  

k
n

k
k
m

k
   

P( E )   
r
r
 
 
n
 
m
Exercise
1
P( E ) | E | 
||
1 2 3 m  r r m 
m n  r  n 
 
  

1 2 3 k  n r k  n k  m  k 
證明

r
r
 
 
n
 m
 m  r  m   n  r  n 
 
  

k
n

k
k
m

k
   

P( E )   
r
r
 
 
n
 
m
Chapter 2
Combinatorial Analysis
Binomial
Coefficients
Binomial Coefficients
 x  y
1
 x  y
2
 x  y
3
 x  y
4
 x  y
0
1
 x y
 x 2  2 xy  y 2
 x 3  3 x 2 y  3xy 2  y 3
 x  4 x y  6 x y  4 xy  y
4
3
2
2
3
4
 x  y
Binomial Coefficients
 x  y
1
 x  y
2
 x  y
3
 x  y
4
 x  y
0
1
 x y
 x 2  2 xy  y 2
 x 3  3 x 2 y  3xy 2  y 3
 x  4 x y  6 x y  4 xy  y
4
3
2
2
3
4
n
?
 x  y
n
?
Binomial Coefficients
 x  y
n
  x  y  x  y  x  y 
x  y
n terms
 x n  ? x n1 y  ? x n2 y 2 
 ? x n k y k 
 yn
 x  y
n
?
Binomial Coefficients
 x  y
n
  x  y  x  y  x  y 
x  y
n boxes
 x n  ? x n1 y  ? x n2 y 2 
n
 
0
n
 
1
n
 
 2
 ? x n k y k 
n
 
k 
 yn
n
 
n
n
Facts: 1.    0 for k  n
k 
n
2.    0 for k  0
k 
Binomial Coefficients
 x  y
n
  x  y  x  y  x  y 
x  y
n
 n  nk k
 n  k nk
   x y    x y
k 0  k 
k 0  k 


 n  nk k
 n  k nk
   x y    x y
k 0  k 
k 0  k 


 n  nk k
 n  k nk
   x y    x y
k   k 
k   k 
n
Properties of Binomial Coefficients
 x  y
n
 n  k nk
   x y
k 0  k 
n
Properties of Binomial Coefficients
 n  n  n  k nk

     1 1
k 0  k 
k 0  k 
n
 1  1
2
n
n
 x  y
n
 n  k nk
   x y
k 0  k 
n
Properties of Binomial Coefficients
n
n
n
k
k nk

   1      1 1
k 0  k 
k 0  k 
n
  1  1
0
n
 x  y
n
 n  k nk
   x y
k 0  k 
n
Properties of Binomial Coefficients
Exercise
Properties of Binomial Coefficients
n不同物件任取k個
第一類取法:
 n  1


k



第二類取法:
 n  1


k

1



 n   n  1   n  1
 


 k   k  1  k 
Properties of Binomial Coefficients
Pascal Triangular
0
 
0
1
 
0
 2
 
0
 3
 
0
 4
 
0
 1
 
 1
 3
 
 2
 3
 
1
 4
 
1
 2
 
 2
 2
 
1
 4
 
 2
 3
 
 3
 4
 
 3
 4
 
 4
 n   n  1   n  1
 


 k   k  1  k 
Properties of Binomial Coefficients
Pascal Triangular
1
0
 
0
1
1
1
 
0
 2
 
0
1
1
 4
 
0
 3
 
0
1
2
4
 3
 
1
1
 2
 
 2
 2
 
1
3
3
 4
 
1
 1
 
 1
6
 4
 
 2
 3
 
 2
1
4
 4
 
 3
 3
 
 3
1
 4
 
 4
Properties of Binomial Coefficients
n
 n  1
k    n

k 
 k  1
吸星大法
Example 7-1
n
 n  k n  n  k nk

x

1


x

x

1


 


k 0  k 
k 0  k 
n
n
n
Fact:     2
k 0  k 
n
Example 7-2
k n n  1
 n  k n  n  k n  n  k n  n  1


k    k    k    n

  n 

k 0  k 
k 1  k  1
k 0  k 
k 1  k 
k 1  k  1 
n
x  n 1 n  1
 n 1 


n 
 n 

x

1

1
x
x 11 
x 0 


x 1 n
k  n 1
n
k 0
n 1 n  1

n

1
n 1

 n 

n
2





k 0  k 
 k 
?
kx+1
簡化版
Example 7-2
k n n  1
n 1 n  1
 n  k n  n 




n 1

n
2
k

n

k

n


    





k
k

1
k
k
k 0  
k 1 
k 1  
k 0 


n
kk+1
簡化版
Example 7-3
k n
 n  k n
n
 n  1
k (k  1)     k (k  1)    n (k  1) 


k
k
k

1
k 0
k 2
  k 2
 


n
n2 n  2
 n  2


n2

n
(
n

1)2
 n(n  1) 
  n(n  1) 

k

2
k
k 2 
k 0 


k n
kk+2
?
Negative Binomial Coefficients
 n 
k  n  k  1
    1 

 k 
 k 
Negative Binomial Coefficients
 n 
k  n  k  1
    1 

 k 
 k 
How to memorize?
 k  11
k (n)
 n 
kk  n
    1 

kk
 k 


Negative Binomial Coefficients
 n 
k  n  k  1
    1 

 k 
 k 
這公式真的對嗎?
k (n+k1) 1 
 n  k  1
k
k
 1 
  1  1 

 k 


k
k
1
 n 
 
 k 
Negative Binomial Coefficients
  n  n(n  1) (n  k  1)
 
k!
 k 
  1
k
n(n  1)
(n  k  1)
k!
(n  k  1) (n  1)n
k  n  k  1
  1
  1 

k!
 k 
k
Chapter 2
Combinatorial Analysis
Some Useful
Mathematic Expansions
Some Useful Mathematic Expansions
Some Useful Mathematic Expansions
1 z z 
1 z 1
1 z
z 2
zz
z2
2
3
z z
z3
2
Some Useful Mathematic Expansions
1
1

1  z 1  ( z )
 1  (  z )  (  z ) 2  (  z )3  (  z ) 4 
 1  z  z 2  z3  z 4 
1
 1 z  z2 
1 z
Some Useful Mathematic Expansions
2
 1 
2
2

1

z

z

1

z

z



 
1 z 
 1  ? z  ? z 2  ? z3  ? z 4 

1
 1 z  z2 
1 z
Some Useful Mathematic Expansions
2

3
2
1

z

z


 1 
2
2

1

z

z

1

z

z



 
1 z 
 1  2 z  3z 2  4 z 3  5 z 4 
 1 
2
2

1

z

z

1

z

z



 
1 z 
 1  ? z  ? z 2  ? z3  ? z 4 
n
 1 
2
3
4

1

?
z

?
z

?
z

?
z



 1 z 

1
 1 z  z2 
1 z
Some Useful Mathematic Expansions
n
n
n
 1 

  1  ( z )    ( z )  1
 1 z 

 n 
 n 
k  nk
   (  z ) 1
   (1)k z k
k 0  k 
k 0  k 


 n  k  1 k
 n  k  1
k
k k
 
(1) (1) z    k z
k 
k 0 
k 0 


Some Useful Mathematic Expansions
Some Useful Mathematic Expansions
z值沒有任何限制
1 z  z2 
 z k  1  z  z 2 
 1  z  z 2 
z
k 1
 z k  2  z k 3 
  z 1  z  z
k 1
2



1
z k 1 1  z k 1



1 z
1 z 1 z
Some Useful Mathematic Expansions
Chapter 2
Combinatorial Analysis
Unordered Samples
with Replacement
Discussion
投返
有序
無序
n
k
?
非投返
n
k
P
n
 
k 
Unordered Samples with Replacement
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
Unordered Samples with Replacement
z
0
 z1  z 2  z 3 
 z
0
 z1  z 2  z 3 

z
0
 z1  z 2  z 3 

此多項式乘開後
zk之係數有何意義?
Unordered Samples with Replacement
z
0
 z
 z1  z 2  z 3 
0
 z1  z 2  z 3 
 1  z  z 2  z 3 
1  z  z
 1  z  z  z 

2
 1 


 1 z 
3
n
 n  k  1 k
 
z
k 
k 0 

n
2
 z3 

 1  z  z
z
2
0
 z1  z 2  z 3 
 z3 


Unordered Samples with Replacement
投返
有序
n
k
非投返
n
k
P
 n  k  1  n 
  
無序 
 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=3
k = 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.
觀察: 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
Example 9
1  z  z
2
 z 3  z 4  z 5  1  z  z 2  z 3  1  z  z 2  z 3 

此多項式乘開後
zk之係數卽為解
Example 9
1  z  z
2
 z 3  z 4  z 5  1  z  z 2  z 3  1  z  z 2  z 3 
1 z6 1 z4 1
 1 



 1  z 6 1  z 4  

1 z 1 z 1 z
1

z


 1 
 1  z 4  z 6  z10  

 1 z 

3
3
 3  j  1 j
 1  z  z  z   
z
j 
j 0 

4
6
10
 j  2 j
 1  z  z  z   
z
j 
j 0 

4
6
10
此多項式乘開後
zk之係數卽為解
Example 9
 j  2 j
1  z  z  z    j z
j 0 


4
6
10


 j  2 j
 j  2 j
 j  2  j 10   j  2  j
4
6
 
z  z  
z  z  
z  z  
z
j 
j 
j 
j 
j 0 
j 0 
j 0 
j 0 

 j  2  j   j  2  j  4   j  2  j 6   j  2  j 10
 
z   
z   
z   
z
j 
j 
j 
j 
j 0 
j 0 
j 0 
j 0 

Coef(zk)=?
此多項式乘開後
zk之係數卽為解
Example 9
 j  2  j   j  2  j  4   j  2  j 6   j  2  j 10


z   
z   
z   
z
j 
j 
j 
j 
j 0 
j 0 
j 0 
j 0 

從z0開始
jk
從z4開始
jk4
從z6開始
jk6
從z10開始
jk10
 k  2 k   k  2 k   k  4 k   k  8  k
 
z   
z   
z   
z
k 0  k
k 4  k  4 
k 6  k  6 
k 10  k  10 


Coef(zk)=?
Example 9
 j  2  j   j  2  j  4   j  2  j 6   j  2  j 10


z   
z   
z   
z
j 
j 
j 
j 
j 0 
j 0 
j 0 
j 0 

從z0開始
jk
從z4開始
jk4
從z6開始
jk6
從z10開始
jk10
 k  2 k   k  2 k   k  4 k   k  8  k
 
z   
z   
z   
z
k 0  k
k 4  k  4 
k 6  k  6 
k 10  k  10 


Coef(zk)=?
Example 9
 k  2 k   k  2 k   k  4 k   k  8  k


z   
z   
z   
z
k 0  k
k 4  k  4 
k 6  k  6 
k 10  k  10 


 k  2 


k



 k  2   k  2 



k
k

4





Coef ( z k )  
 k  2    k  2    k  4 
 k   k  4   k  6 

 k  2   k  2   k  4   k  8 





k
k

4
k

6
 
 
  k  10 

0k 4
4k 6
6  k  10
10  k
Example 9
 k  2  k  2  k  4  k  8 
Coef ( z k )  




k
k

4
k

6

 
 
  k  10 
 k  2 


k



 k  2   k  2 



k
k

4





Coef ( z k )  
 k  2    k  2    k  4 
 k   k  4   k  6 

 k  2   k  2   k  4   k  8 





k
k

4
k

6
 
 
  k  10 

0k 4
4k 6
6  k  10
10  k
Chapter 2
Combinatorial Analysis
Derangement
Derangement
最後! ! !
每一個人都拿
到別人的帽子
錯排
1. P( E0n )  ?
2. lim P( E0n )  ?
Example 10
n
3. P ( Ekn )  ?
n
k
E
n人中正好k人拿
對自己的帽子
n
0
E
n人中無人拿
對自己的帽子
1. P( E0n )  ?
2. lim P( E0n )  ?
Example 10
n
3. P ( Ekn )  ?
n
k
E
 ?
E ?
n
k
n
0
E
E ?
n
0
1. P( E0n )  ?
2. lim P( E0n )  ?
Example 10
n
k
n
0
E
E
n
3. P ( Ekn )  ?
 n人中正好k人拿對自己的帽子
  n!
 n人中無人拿對自己的帽子
P ( E )  ?1/2!
2
0
P ( E )  ?2/3!
3
0
2
3
1
2
1
3
1
2
1
2
1
2
3
1. P( E0n )  ?
Example 10
n
k
n
0
E
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
2. lim P( E0n )  ?
n
3. P ( Ekn )  ?
  n!
令Ai表第i個人拿了自己帽子
P( E )  1  P( E )  1  P( A1  A2 
n
0
n
0
 An )
1. P( E0n )  ?
Example 10
Ai表第i個人拿了自己帽子
P( E0n )  1  P( A1  A2 
 An )
2. lim P( E0n )  ?
n
3. P ( Ekn )  ?
1. P( E0n )  ?
(n  1)!
n!
Example 10
P ( Ai ) 
2. lim P( E0n )  ?
n
3. P ( Ekn )  ?
Ai表第i個人拿了自己帽子
P( E0n )  1  P( A1  A2 
 An )
...
1
...
1
2
n
1. P( E0n )  ?
(n  1)!
n!
(n  2)!
,i  j
n!
Example 10
P ( Ai ) 
P ( Ai  A j ) 
2. lim P( E0n )  ?
n
3. P ( Ekn )  ?
Ai表第i個人拿了自己帽子
P( E0n )  1  P( A1  A2 
 An )
...
1
2
...
1
2
n
n
計  項
1
(n  1)!
n!
n
計  項
 2
1. P( E0n )  ?
(n  2)!
,i  j
n!
Example 10
P ( Ai ) 
P ( Ai  A j ) 
Ai表第i個人拿了自己帽子
P( E0n )  1  P( A1  A2 
2. lim P( E0n )  ?
 An )
n
3. P ( Ekn )  ?
 n  (n  k )!
Sk   
 k  n!
n!
(n  k )!

k !(n  k )! n !

1
k!
(n  3)!
P( Ai  Aj  Ak ) 
,i  j  k
n!
n
計  項
 3
1. P( E0n )  ?
Example 10
2. lim P( E0n )  ?
n
3. P ( Ekn )  ?
Ai表第i個人拿了自己帽子
P( E0n )  1  P( A1  A2 
 An )
 1 1
P( E )  1  1   
 2! 3!
  1
n
0
n 1
 n  (n  k )!
Sk   
 k  n!
n!
(n  k )!

k !(n  k )! n !
1
n !

 1
1
k!
1 1
 
2! 3!
  1
n 1
1
n!
1. P( E0n )  ?
Example 10
2. lim P( E0n )  ?
n
3. P ( Ekn )  ?
Ai表第i個人拿了自己帽子
P( E0n )  1  P( A1  A2 
 An )
 1 1
P( E )  1  1   
 2! 3!
  1
n
0
 1 1
P ( E )  1  1   
 2! 2
2
0
1
n !
 1 1 1
P( E )  1  1    
 2! 3! 3
 1 1 1 3
P( E )  1  1     
 2! 3! 4! 8
4
0
n 1
3
0
1. P( E0n )  ?
2. lim P( E0n )  ?
Example 10
n
3. P ( Ekn )  ?
Ai表第i個人拿了自己帽子
P( E0n )  1  P( A1  A2 
 An )
n 1 1 
 1 1
P( E )  1  1      1
n !
 2! 3!
n 1 1 
1 1 1
 1        1
n !
1! 2! 3!
n
0
1 1 1
P( E )  1    
1! 2! 3!
n
0
1
  1
n!
n
1. P( E0n )  ?
2. lim P( E0n )  ?
Example 10
n
3. P ( Ekn )  ?
Ai表第個人拿了自己帽子
P( E0n )  1  P( A1  A2 
 An )
n 1 1 
 1 1
P( E )  1  1      1
n !
 2! 3!
n 1 1 
1 1 1
 1        1
n !
1! 2! 3!
n
0
lim P( E )  e
n
0
n
1 1 1
P( E )  1    
1! 2! 3!
n
0
1
  1
n!
n
1
1. P( E0n )  ?
2. lim P( E0n )  ?
Example 10
P( E ) 
n
0
P( E ) 
n
k
E0n
n!
Ekn
n!
1 1 1
P( E )  1    
1! 2! 3!
n
0
n
3. P ( Ekn )  ?
E0n  n ! P ( E0n )
Ekn  ?
1
  1
n!
n
1. P( E0n )  ?
2. lim P( E0n )  ?
Example 10
P( E ) 
n
0
E0n
n!
nk
E
n 0
n
P( Ek ) 
 
n!  k  n!
nk
E
 n  (n  k )! 0
 
 k  n ! (n  k )!
Ekn
1
P( E )  P( E0n  k )
k!
n
k
n
3. P ( Ekn )  ?
E0n  n ! P ( E0n )
n
E ? 
k 
n
k

E0n  k
...
...
...
...
k matches
nk mismatches
1. P( E0n )  ?
Example 10
2. lim P( E0n )  ?
n
3. P ( Ekn )  ?
1
1 1
3
P ( E )  ? P  E0   
2!
2! 3
5
2
1
P( E )  P( E0n  k )
k!
n
k
Remark
n
P
(
E
)

?
1

k 0
n
k
Chapter 2
Combinatorial Analysis
Calculus
Some Important Derivatives



Derivatives for
multiplications —
duv
 u v  uv
dx
Derivatives for
divisions —
d  v  uv  u v
 
2
dx  u 
u
Chain rule —
dy dy du


dx du dx
y  f (u )
u  g ( x)
L’Hopital rule
f ( x) 0

Suppose as x  c, we have
 or .
g ( x) 0

f ( x)
f ( x)
lim
 lim
x c g ( x )
x c g ( x)
Examples
Integration by Part
duv  udv  vdu
duv

udv

vdu



uv   udv   vdu
Integration by Part
b
b

udv

uv

vdu


uv   udv   vdu
a
udv  uv a   vdu
b
a
The Gamma Function

( )   x
0
e dx,   0
 1  x

( )   x
0
Example 12
e dx,   0
 1  x

( )   x
0
e dx,   0
 1  x
Example 12

(1)   e x dx
0
 e
x 
0
 0  (1)
1

( )   x
0
Example 12
e dx,   0
 1  x

( )   x
e dx,   0
 1  x
0
Example 12


( )   x e dx   x 1de x
 1  x
0
0

 x
 1  x 
 x
e
0
 1  x 
e
0
0

  e x dx 1
0

 (  1)  x
 (  1)(  1)
0

 2  x
e dx
(1)


( )   x
0
Example 12
e dx,   0
 1  x
( )  (  1)(  1)
 (  1)(  2)(  2)
 (  1)(  2)(  3)
 (  1)!
(1)

( )   x
0
Example 12
e dx,   0
 1  x



e
 x2 / 2
dx  ?

( )   x
e dx,   0
 1  x
0

Example 12



e
 x2 / 2
dx 
I 


2
2




 
0
0
e



e
x2  y 2

2
 x2 / 2



dx  e
 y2 / 2

dxdy  
2
0


0
e
 x2 / 2
dx  ?
dy  I  2
e
2
r2 / 2
rdrd
2

2
r 
r / 2
e
d   d   d  2
0
2
2

( )   x
e dx,   0
 1  x
0
Example 12



e
 x2 / 2
dx  2

( )   x
e dx,   0
 1  x
0
Example 12



e
 x2 / 2
dx  2
x 

1
1/ 2  x
     x e dx   x1/ 2e x dx
x 0
2 0
2 1/ 2
2
y2

2




y
y
2
y /2
  y2   e
d 

0
2
 2 
 2 

y 
y 0

2  y2 / 2
 y2 / 2
e
 ydy  2  e
dy
0
y
2   y2 / 2

e
dy  

2 
Let x  y 2 / 2

( )   x
0
Example 12
e dx,   0
 1  x

( )   x
0
e dx,   0
 1  x
Example 12

3 1 1
   ?   
2
2 2 2
5 3 3  3 
   ?  
4
2 2 2
 7  5  5  15 
   ?   
8
2 2 2