1. Discrete Mathematics
Chapter 5 Counting

2. §5.1 The Basics of counting
 A counting problem: (Example 15)
Each user on a computer system has a password, which is six to eight characters long, where each characters is an uppercase letter or a digit. Each password must contain at least one digit. How many possible passwords are there?
 This section introduces
a variety of other counting problems
the basic techniques of counting.

3. Basic counting principles
 The sum rule:
If a first task can be done in n1 ways and
a second task in n2 ways, and if these tasks
cannot be done at the same time. then there
are n1+n2 ways to do either task.
Example 11
Suppose that either a member of faculty or a student is chosen as a representative to a university committee. How many different choices are there for this representative if there are 37 members of the faculty and 83 students? n1 n2
n1 + n2 ways

4. 23, 15 and 19 possible projects respectively.
Example A student can choose a computer project from one of three lists. The three lists contain
23, 15 and 19 possible projects respectively.
How many possible projects are there to choose from?
Sol: =57 projects.
 The product rule:
Suppose that a procedure can be broken down
into two tasks. If there are n1 ways to do the
first task and n2 ways to do the second task after
the first task has been done, then there are
n1 n2 ways to do the procedure. n1 n2
n1 × n2
ways

5. Example 2 The chair of an auditorium (大礼堂) is
　to be labeled with a letter and a positive integer
　not exceeding 100. What is the largest number of
　chairs that can be labeled differently?
Sol: 26 × 100 = ways to label chairs.
letter
Example 4 How many different bit strings are there of length seven?
Sol:　1　2　 3　 4　 5　 6 　7
□ □ □ □ □ □ □
↑ ↑ ↑ ↑ ↑ ↑ ↑
0,1 0,1 0, ,1
→ 27 种

6. Example 6 How many functions are there from a
How many different license plates (车牌) are available if each plate contains a sequence of 3 letters followed by 3 digits ?
Sol: □ □ □ □ □ □ →263．103
letter digit
Example 6 How many functions are there from a
set with m elements to one with n elements?
Sol: f(a1)=? 可以是b1～ bn, 共 n种
f(a2)=? 可以是b1～ bn, 共 n种 ：
f(am)=? 可以是b1～ bn, 共 n种
∴nm a1 a2 ． am b1 b2 bn f

7. Example 7 How many one-to-one functions are there
from a set with m elements to one with n element? (m  n)
Sol: f(a1) = ? 可以是b1～ bn, 共 n 种
f(a2) = ? 可以是b1～ bn, 但不能= f(a1), 共 n-1 种
f(a3) = ? 可以是b1～ bn, 但不能= f(a1), 也不能=f(a2), 共 n-2 种 ：
f(am) = ? 不可=f(a1), f(a2), ... , f(am-1), 故共n-(m-1)种
∴共 n．(n-1)．(n-2)．...．(n-m+1)种 1-1 function #

8. Example 15 Each user on a computer system has a password which is 6 to 8 characters long, where each character is an uppercase letter or a digit. Each password must contain at least one digit. How many possible passwords are there?
Sol: Pi : # of possible passwords of length i , i=6,7,8
P6 =
P7 =
P8 =
∴ P6 + P7 + P8 = 种

9. Example 14
In a version of Basic, the name of a variable
is a string of one or two alphanumeric characters, where
uppercase and lowercase letters are not distinguished.
Moreover, a variable name must begin with a letter and
must be different from the five strings of two characters
that are reserved for programming use. How many different variable names are there in this version of Basic?
Sol:
Let Vi be the number of variable names of length i.
V1 =26
V2 =26．36 – 5
∴ ．36 – 5 different names.

10. ※ The Inclusion-Exclusion Principle (排容原理)
A B
Example How many bit strings of length eight either start with a 1 bit or end with the two bits 00 ?
Sol:
１ ２ ３ ４ ５ ６ ７ ８
□ □ □ □ □ □ □ □
↑ ↑
① 1 0,1 0, → 共27种
② 　 → 共26种
③ → 共25种
 种

11. ※ Tree Diagrams
Example How many bit strings of length four do not have two consecutive 1s ?
Sol:
Exercise: 11, 17, 23, 27, 38, 39, 47, 53 1
(0000)
(0001)
(0010)
(0100)
(0101)
(1000)
(1001)
(1010)
∴ 8 bit strings 1
bit 3 1 1
bit 1

12. Ex 38. How many subsets of a set with 100 elements
have more than one element ?
Sol:
Ex A palindrome (回文) is a string whose reversal
is identical to the string. How many bit strings
of length n are palindromes ?
( abcdcba 是回文, abcd 不是 )
Sol: If a1a2 ... an is a palindrome, then a1=an, a2=an-1, a3=an-2, …
Thm. 4 of §4.3
空集合及 只有1个元素的集合 放 不放 放 不放 放 不放

13. §5.2 The Pigeonhole Principle (鸽笼原理)
Theorem 1 (The Pigeonhole Principle)
If k+1 or more objects are placed into k boxes, then there is at least one box containing two or more of the objects.
Proof
Suppose that none of the k boxes contains more than one object. Then the total number of objects would
be at most k. This is a contradiction.
Example Among any 367 people, there must be
at least two with the same birthday, because there
are only 366 possible birthdays.

14. Example 2 In any group of 27 English words, there must be at least two that begin with the same letter.
Example 3 How many students must be in a class to
guarantee that at least two students receive the
same score on the final exam ? (0~100 points)
Sol: (101+1)
Theorem 2. (The generalized pigeon hole principle)
If N objects are placed into k boxes, then
there is at least one box containing at least
objects.
e.g. 21 objects, 10 boxes  there must be one box containing at least objects.

15. Example 5 Among 100 people there are at least
who were born in the same month. ( 100 objects, boxes )

16. Example 10 During a month with 30 days a baseball
(跳过)
Example During a month with 30 days a baseball
team plays at least 1 game a day, but no more
than 45 games. Show that there must be a period
of some number of consecutive days during which the
team must play exactly 14 games.
day 1 2 3 4 5 30
# of game
存在一段时间的game数和=14

17. Let aj be the number of games played on or before the
(跳过)
Sol:
Let aj be the number of games played on or before the
jth day of the month. (第1天～第j天的比赛数和)
Then is an increasing sequence of distinct integers with
Moreover, is also an increasing
sequence of distinct integers with
There are 60 positive integers
between 1 and 59. Hence, such that

18. Def. Suppose that is a sequence of numbers.
A subsequence of this sequence is a sequence of the form where
e.g sequence: 8, 11, 9, 1, 4, 6, 12, 10, 5, 7
subsequence: 8, 9, ()
9, 11, 4, 6 ()
Def. A sequence is called increasing (递增) if
A sequence is called decreasing (递减) if
A sequence is called strictly increasing (严格递增) if
A sequence is called strictly decreasing (严格递减) if

19. (跳过)
Theorem 3. Every sequence of n2+1 distinct real numbers contains a subsequence of length n+1 that is either strictly increasing or strictly decreasing.
Example The sequence 8, 11, 9, 1, 4, 6, 12, 10, 5, 7 contains 10=32+1 terms (i.e., n=3). There is a strictly increasing subsequence of length four, namely, 1, 4, 5, 7. There is also a decreasing subsequence of length 4, namely, 11, 9, 6, 5.
Exercise Construct a sequence of 16 positive integers that has no increasing or decreasing subsequence of 5 terms.
Sol:
Exercise: 5, 13, 15, 31

20. §5.3 Permutations(排列) and Combinations(组合)
Def. A permutation of a set of distinct objects is an ordered arrangement of these objects. An ordered arrangement of r elements of a set is called an r-permutation.
Example 2. Let S = {1, 2, 3}. The arrangement 3,1,2 is a permutation of S. The arrangement 3,2 is a 2-permutation of S.
Theorem The number of r-permutation of a set with n
distinct elements is
位置： … r
　　　□ □ □ … □
放法： …

21. Example 4. How many different ways are there to select
a first-prize winner (第一名), a second-prize winner, and a third-prize winner from 100 different people who have entered a contest ?
Sol:
Example 6. Suppose that a saleswoman has to visit
8 different cities. She must begin her trip in a specified city, but she can visit the other cities in any order she wishes. How many possible orders can the saleswoman use when visiting these cities ?

22. Def. An r-combination of elements of a set is an
unordered selection of r elements from the set.
Example 9 Let S be the set {1, 2, 3, 4} Then {1, 3, 4} is a 3-combination from S.
Theorem 2 The number of r-combinations of a set with n elements, where n is a positive integer and r is
an integer with , equals
pf :
称为 binomial coefficient

23. Example 10. We see that C(4,2)=6, since the
2-combinations of {a,b,c,d} are the six subsets
{a,b}, {a,c}, {a,d}, {b,c}, {b,d} and {c,d}
Corollary 2. Let n and r be nonnegative integers with r  n.
Then C(n,r) = C(n,n-r)
pf : From Thm 2.
组合意义：选 r 个拿走,相当于是选 n - r 个留下.

24. Example How many ways are there to select 5 players from a 10-member tennis team to make a trip to a match at another school ?
Sol: C(10,5)=252
Example 15. Suppose there are 9 faculty members in the math department and 11 in the computer science department. How many ways are there to select a committee if the committee is to consist of 3 faculty members from the math department and 4 from the computer science department?
Sol:
Exercise: 3, 11, 13, 21, 33, 34.

25. §5.4 Binomial Coefficients (二项式系数)
Example 1.
要产生 xy2 项时， 需从三个括号中选两个括号提供 y，剩下一个则提供 x
(注意：同一个括号中的 x 跟 y 不可能相乘)
∴共有 种不同来源的 xy2 ∴
 xy2 的系数 =
Theorem (The Binomial Theorem, 二项式定理)
Let x,y be variables, and let n be a positive integer, then

26. Example 4. What is the coefficient of x12y13 in
the expansion of ?
Sol: ∴
Cor Let n be a positive integer. Then
pf : By Thm 1, let x = y = 1
Cor Let n be a positive integer. Then
pf : by Thm 1. (1-1)n = 0

27. Theorem 2. (Pascal’s identity)
Let n and k be positive integers with n  k
Then
PASCAL’s triangle … …

28. pf : ①(algebraic proof, 代数证明)
②(combinatorial proof, 组合意义证明): n ‧ 1 k 1 k n ‧ 1
k-1 + n ‧
取法 = 

29. Theorem 3. (Vandermode’s Identity)
pf : m n r
m n m n m n
↓ ↓ ↓ ↓ ↓ ↓
r, r-1, , r =
C(m+n, r) =

30. Ex 33. Here we will count the number of paths
between the origin (0,0) and point (m,n) such
that each path is made up of a series of
steps, where each step is a move one unit
to the right or a more one unit upward.
(5,3)
(0,0)
Each path can be represented by a bit string consisting of m 0s and n 1s.
(→) (↑)
Red path对应的字符串:
There are paths of the desired type.
Exercise: 7, 21, 28

31. §5.5 Generalized Permutations and Combinations
Permutations with Repetition
Example 1 How many strings of length r can be formed from the English alphabet?
Sol. 26r
Thm 1. The number of r-permutations of a set of n objects with repetition allowed is n r.

32. Combinations with Repetition
Example 3 How many ways are there to select five bills from a cash box containing $1 bills, $2 bills, $5 bills, $10 bills, $20 bills, $50 bills, and $100 bills? (假设顺序不计、同面额不区分、每种面额至少五张)
Sol.
的个数 表示张数
隔开的bar 面额 1 2 5 10 20 50
100 张数 3
 可用字符串 |  |  | |  |  | 代表
 可用字符串   |  | |   | | | 代表
 可用字符串    | | |  | | |  代表 …
7种面额需6个bar隔开，故6个bar 跟5张bill组合  C(11, 5)

33. Thm 2. There are C(r+n-1, r) r-combinations from
a set with n elements when repetition of elements is
allowed.
pf : (视为有 r 个＊，要放入 n 个区域， 故需插入 n-1个 bar 将这些＊隔成n 区)
例： 设n = 4, 集合为 , r = 6
a1出现2次
a2出现1次
a3不出现
a4出现3次 取法
＊＊｜＊｜｜＊＊＊
∴ 种方法 #

34. Example 4. Suppose that a cookie shop has 4 different
kinds of cookies. How many different ways can
6 cookies be chosen?
Sol: 6个cookie插入3个bar 种
Example How many solutions does the equation
x1 + x2 + x3 = 11
have, where x1, x2, x3 are nonnegative integers?’
Sol: 11个1间要插入2个bar 种

35. 可改成 (x1 -1) + (x2 -2) + (x3 -3) = 11 - 1 - 2 - 3 = 5
此时相当于是 y1 + y2 + y3 = 5
其中 y1 = x1 -1, y2 = x2 -2, y3 = x3 -3 且 y1, y2 , y3 N
∴5个1间要插入2个bar  种
(注意： case y1 = 1, y2 = 3, y3 = 1 相当于 x1 =2, x2 =5, x3 =4)
※若上题再改为 1 x1  3, x2  2, x3  3，则需排除 x1 > 3的情形 (即x1  4的情形)
因 (x1 -4) + (x2 -2) + (x3 -3) = = 2
∴共有 种
Exercise: 15, 20, 25

36. ※ Permutations with indistinguishable objects
Example 7. How many different strings can be made
by reordering the letters of the word SUCCESS ?
Sol:
有3个S, 2个C, 1个U 及 1个E,
可放S的位置有 种
剩下的4个位置中可放C的有 种
剩下的2个位置中可放U的有 种
剩下的1个位置中可放E的有 种
∴共 种

37. Thm 3. The number of different permutations of n objects,
where type 1： n1 种
type 2： n2 种 is
type k： nk 种
pf :
Exercise: 17, 31, 36, 65

38. Distributing objects into Boxes
Distinguishable Objects and distinguishable boxes
Example 8. How many ways are there to distribute
hands of 5 cards to each of four players from the
standard deck of 52 cards?
Sol: player 1： 种
player 2： 从剩下的牌再发5张 
注意：上题相当于将52张牌放进5个不同的box的放法,
即 box 1 的给 player 1
　　　　 box 2 的给 player 2
　　　　 box 3 的给 player 3
　　　　 box 4 的给 player 4
　　　而 box 5 的是剩下的牌.
Exercise: 45

39. Thm 4. The number of ways to distribute n distinguishable
objects into k distinguishable boxes so that ni
objects are placed into box i, i=1, 2, …, k, equals
(跟Thm 3相等)
Indistinguishable Objects and distinguishable boxes
Example 9. How many ways are there to place 10 indistinguishable balls into eight distinguishable bins?
Sol: C(8+10-1, 10)
There are C(n+r-1, n-1) ways to place r indistinguishable objects into n distinguishable boxes.

40. Distinguishable Objects and indistinguishable boxes
Example 10. How many ways are there to put four different employees into three indistinguishable offices, when each office can contain any number of employees?
Sol: employees: A, B, C, D
4人在同一office: {{A, B, C, D}}  1种
3人在同一office, 1人在另一office: {{A, B, C}, {D}}, {{A, B, D}, {C}}, …  4种
2人在同一office, 2人在另一office: {{A, B}, {C, D}}, {{A, D}, {B, C}}, …  3种
2人在同一office,另2人分别在剩下两个office: {{A, B}, {C}, {D}}, {{A, D}, {B}, {C}}, …  6种
共 14种
Note. There is no simple closed formula. 可参考 Stirling numbers of the second kind. (p. 378)

41. Indistinguishable Objects and indistinguishable boxes
Example 11. How many ways are there to pack six copies of the same book into four identical boxes, where a box can obtain as many as six books?
Sol:
3, 3
2, 2, 1, 1 6
3, 2, 1
5, 1
共 9种
3, 1, 1, 1
4, 2
2, 2, 2
4, 1, 1
Note. This problem is the same as writing n as the sum of at most k positive integers in nonincreasing order. That is, a1+a2+…+aj = n, where a1, a2, …, aj are positive integers with a1  a2  …  aj and j  k.
No simple closed formula exists.