Discrete Mathematicsand Its Applications

Sixth EditionBy Kenneth Rosen

Chapter 5 Counting

歐亞書局

5.1 The Basics of Counting5.2 The Pigeonhole Principle5.3 Permutations and

Combinations5.4 Binomial Permutations and

Combinations5.5 Generalized Permutations and

Combinations 5.6 Generating Permutations and

Combinations

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5.1 The Basics of Counting

• Basic Counting Principles – The product rule: Suppose that a procedure can

be broken down into a sequence of two tasks. If there are n1 ways to do the first task and for each of these ways, there are n2 ways to do the second task, then there are n1n2 ways to do the procedure

• Ex.1-10

– The sum rule: If a task can be done either in one of n1 ways or in one of n2 ways, where none of the set of n1 ways is the same as any of the set of n2 ways, then there are n1+n2 ways to do the task

• Ex. 11-13

• The product rule: If A1,A2 ,…,Am are finite sets, then the number of elements in the Cartesian product of these sets is the product of the number of elements in each set.– |A1 A2 … Am|= |A1||A2| …|Am|

• The sum rule: If A1,A2 ,…,Am are disjoint finite sets, then the number of elements in the union of these sets is the sum of the number of elements in each set.– |A1 A2 … Am|= |A1|+|A2|+ …+|Am|

More Complex Counting Problems

• Ex. 14-16

FIGURE 1 (5.1)

FIGURE 1 Internet Addresses (IPv4).

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The Inclusion-Exclusion Principle

• Suppose a task can be done in n1 or n2 ways, but some of the set of n1 ways are the same as some of the n2 ways, we have to subtract the number of ways to do the task that is both among the set of n1 ways and the set of n2 ways– The subtraction principle– |A1 A2 |= |A1|+|A2|- |A1 A2 |– Ex. 17-18

Tree Diagrams

• Counting problems can be solved using tree diagrams– Leaves: possible outcomes– Ex. 19-21

FIGURE 2 (5.1)

FIGURE 2 Bit Strings of Length Four without Consecutive 1s.

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FIGURE 3 (5.1)

FIGURE 3 Best Three Games Out of Five Playoffs.

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FIGURE 4 (5.1)

FIGURE 4 Counting Varieties of T-Shirts.

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5.2 The Pigeonhole Principle

• If there are more pigeons than pigeonholes, then there must be at least one pigeonhole with at least two pigeons in it

• Theorem 1: (The Pigeonhole Principle) If k is a positive integer and k+1 or more objects are placed into k boxes, then there is at least one box containing two or more of the objects.– Also called the Dirichlet drawer principle– Proof (by contraposition)

FIGURE 1 (5.2)

FIGURE 1 There Are More Pigeons Than Pigeonholes.

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• Corollary 1: A function from a set with k+1 or more elements to a set with k elements is not one-to-one.– proof

• Ex.1-4

The Generalized Pigeonhole Principle

• Theorem 2: (The Generalized Pigeonhole Principle) If N objects are placed into k boxes, then there is at least one box containing at least N/k objects.– Proof– Ex. 5-8

Some Elegant Applications of the Pigeonhole Principle

• Ex. 10• Ex. 11• Suppose a1, a2, …, aN is a sequence

of real numbers, a subsequence of this sequence is a sequence of the form ai1, ai2, …aim, where 1<=i1<i2<…<im<=N

• Theorem 3: Every sequence of n2+1 distinct real number contains a subsequence of length n+1 that is either strictly increasing or strictly decreasing.– Ex.12– Ramsey theory

• Ex.13• Ramsey number R(m,n): the minimum number

of people at a party such that there are either m mutual friends or n mutual enemies

5.3 Permutations and Combinations

• Permutations– A permutation of a set of distinct objects

is an ordered arrangement of these objects

– r-permutation: an ordered arrangement of r elements of a set

– Ex.1– Ex.2– P(n,r): the number of r-permutations of a

set with n elements

• Theorem 1: If n is a positive integer and r is an integer with 1<=r<=n, then there are P(n,r)=n(n-1)(n-2)…(n-r+1) r-permutations of a set with n distinct elements.– P(n,0)=1– P(n,n)=n!

• Corollary 1: If n and r are integers with 0<=r<=n, then P(n,r)=n!/(n-r)!– Ex. 4-7

• Combinations– Finding the number of subsets of a

particular size– r-combination: an unordered selection of

r elements from the set– Ex.8-9– C(n,r): the number of r-combinations of

a set with n elements• Also denoted by , and is called a binomial

coefficient

r

n

• Theorem 2: The number of r-combinations of a set with n elements, where n is a nonnegative integer and r is an integer with 0<=r<=n, equalsC(n,r)=n!/(r!(n-r)!)– Proof– C(n,r)=n(n-1)…(n-r+1)/r!

• Corollary 2: Let n and r be nonnegative integers with r<=n. Then C(n,r)=C(n,n-r).– Proof

• Definition 1: A combinatorial proof of an identity: using counting arguments to prove that both sides of the identity count the same objects but in different ways.– Ex.12-15

5.4 Binomial Coefficients

• (x+y)n

– Ex.1

• Theorem 1: (The Binomial Theorem) Let x and y be variables, and let n be a nonnegative integer. Then,(x+y)n=j=0..n C(n,r)xn-jyj

=C(n,0)xn+C(n,1)xn-1y+…+C(n,n-1)xyn-

1+C(n,n)yn

– Proof (combinatorial proof)– Ex. 2-4

• Corollary 1: Let n be a nonnegative integer. Then,k=0..nC(n,k)=2n.– Proof

• Corollary 2: Let n be a positive integer. Then,k=0..n (-1)kC(n,k)=0.– Proof– C(n,0)+C(n,2)+C(n,4)+…=C(n,1)+C(n,3)+C(n,5)+…

• Corollary 3: Let n be a nonnegative integer. Then,k=0..n 2k

C(n,k)=3n.– Proof

Pascal’s Identity and Triangle

• Theorem 2: (Pascal’s Identity) Let n and k be positive integers with n>=k. Then,C(n+1,k)=C(n,k-1)+C(n,k).– Proof– This can be used to recursively define

binomial coefficients.

• Pascal’s triangle– C(n,k), k=0, 1, …, n

FIGURE 1 (5.4)

FIGURE 1 Pascal’s Triangle. 歐亞書局 P. 367

Some Other Identities of the Binomial Coefficients

• Theorem 3: (Vandermonde’s Identity) Let m, n, and r be nonnegative integers with r not exceeding either m or n. Then,C(m+n, r)=k=0..rC(m,r-k)C(n,k).– Proof

• Corollary 4: If n is a nonnegative integer, then,C(2n,n)=k=0..nC(n,k)2

– Proof• Theorem 4: Let n and r be nonnegative

integers with r<=n. Then, C(n+1,r+1)= j=r..nC(j,r).– Proof

5.5 Generalized Permutations and Combinations

• Permutations with repetition– Theorem 1: The number of r-permutations of a

set of n objects with repetition allowed is nr.– Ex.1

• Combinations with repetition– Theorem 2: There are C(n+r-1,r)=C(n+r-1,n-1)

r-combinations from a set with n elements when repetition of elements is allowed.

– Ex.2-3– Ex.4-6

FIGURE 1 (5.5)

FIGURE 1 Cash Box with Seven Types of Bills.

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FIGURE 2 (5.5)

FIGURE 2 Examples of Ways to Select Five Bills. 歐亞書局 P. 372

TABLE 1 (5.5)

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• Permutations with indistinguishable objects– Theorem 3: The number of different

permutations of n objects, where there are n1 indistinguishable objects of type 1, n2 indistinguishable objects of type 2, …, and nk indistinguishable objects of type k, isn!/(n1!n2!...nk!).

– Ex.7

• Distributing objects into boxes– Objects: distinguishable/indistinguishable– Boxes: distinguishable/indistinguishable

• Distinguishable objects and distinguishable boxes– Ex.8– Theorem 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, equalsn!/(n1!n2!...nk!).

• Indistinguishable objects and distinguishable boxes– The same as counting the number of n-

combinations for a set with k elements when repetitions are allowed• Ex.9

• Distinguishable objects and indistinguishable boxes– More difficult, no simple closed formula– Ex.10

• Indistinguishable objects and indistinguishable boxes– No simple closed formula– Ex.11– The same as partitioning positive

integer n into k positive integers.

5.6 Generating Permutations and Combinations

• Generating permutations– Lexicographic (dictionary) ordering

• Permutation a1a2…an precedes b1b2…bn, if for some k, with 1<=k<=n, a1=b1, a2=b2, …, ak-1=bk-1, and ak<bk.

• Ex.1

• For a1a2…ajaj+1…an such thataj<aj+1

aj+1>aj+2>…>an

– Find the smallest among aj+1, …, an that is > aj

– Increasing order for the remaining numbers

• Ex.2• Ex.3

• Algorithm 1: Generating the next permutation in lexicographic order– Procedure next_permutation(a1a2…an)

j:=n-1while aj>aj+1 j:=j-1k:=nwhile aj>ak k:=k-1interchange aj and akr:=ns:=j+1while r>sbegin interchange ar and as r:=r-1 s:=s+1end

• Generating Combinations– Correspondence with bit strings of length n– Binary expansion of an integer between 0

and 2n-1– At each stage, the next binary expansion is

found by locating the first position from the right that is not a 1, then changing all the 1s to the right of this position to 0s and making this first 0 a 1.• Ex.4

• Algorithm 2: Generating the next larger bit string– Procedure next_bit_string(bn-1bn-2…b1b0)

i:=0while bi=1begin bi:=0 i:=i+1end

• The r-combinations can be listed using lexicographic order on the sequence– The next combinations after a1a2…ar can

be obtained by:• Locate the last ai such that ai!=n-r+i

• Replace ai with ai+1, aj with ai+j-i+1, for j=i+1, …, r

• Ex.5

• Algorithm 3: generating the next r-combination in lexicographic order– Procedure next_r_comb(a1a2…ar)

i:=rwhile ai=n-r+i i:=i-1ai:=ai+1for j:=i+1 to r aj:=ai+j-i

Thanks for Your Attention!