4.2 Permutations of sets

4.2.1 Basic counting principles Theorem 4.3(Addition principle): If A1, A2, …

, An are disjoint sets, then the number of elements in the union of these sets is the sum of the numbers of elements in them.

| A1 A∪ 2 … A∪ ∪ n |=|A1|+|A2|+…+|An|

Example1: A student wishes to take either a mathematics course or biology course, but not both. If there are 4 mathematics course and 2 biology course for which the student has the necessary prerequisites, then the student can choose a course to take in 4+2=6 ways.

Theorem 4.4(Multiplication principle): Let A and B be two finite sets. Let |A|=p and |B|=q, then |A×B|=p×q.

Example2: A student wishes to take a mathematics course and a biology course. If there are 4 mathematics course and 2 biology course for which the student has the necessary prerequisites, then the student can choose courses to take in 4×2=8 ways.

4.2.2 Permutations of sets

An ordered arrangement of r elements of an n-element set is called an r-permutation.

We denote by p(n,r) the number of r-permutations of an n-element set. If r>n, then p(n,r)=0. An n-permutation of an n-element set S is called a permutation of S. A permutation of a set S is a listing of the elements of S in some order.

Theorem 4.5: For n and r positive integers with rn, p(n,r)=n(n-1)…(n-r+1) Proof:In constructing an r-permutation of an n-

element set, we can choose the first item in n ways, the second item in n-1 ways whatever choice of the first item,… , and the rth item in n-(r-1) ways whatever choice of the first r-1 items. By the multiplication principle the r items can be chosen in n(n-1)…(n-r+1) ways.

We define n! by n!= n(n-1)…2•1 with the convention that 0!=1.Thus p(n,r)=n!/(n-r)!.

Example:What is the number of ways to order the 26 letters of the alphabet so that no two of the vowels a,e,i,o,and u occur consecutively?(元音字母中任意两个都不得相继出现 )

Solution:The first task is to decide how to order the consonants among themselves.

21! Our second task is put the vowels in these places. p(22,5)=22!/17! By the multiplication principle, the numble of

ordered arrangements of the letters of the alphabet with no two vowels consecutive is 21!22!/17! .

Example: What is the number of ways to order the 26 letters of the alphabet so that it contains exactly seven letters between a and b?

a…….b ， P(24,7) b…….a ， P(24,7) between a and b 2P(24,7) P(18,18)=18!. 2P(24,7)18! linar permutation circular permutation

linar permutation circular permutation linar permutation 12345 linar permutation 45123 circular permutation

For example, the circular permutation

arises from each of the linear permutation

12345 23451 34512 45123 51234

Theorem 4.6: The number of circular r-permutations of a set of n elements is given by p(n,r)/r=n!/r(n-r)! . In particular,the number of circular permutations of n elements is (n-1)! .

Proof: The set of linear r-permutations can be partitioned into parts in such a way that two linear r-permutations are in the same part if only if they correspond to the same circular r-permutations .

Thus the number of circular r-permutations equals the number of parts.

Since each part contains r linear r-permutations, the number of parts is the number p(n,r) of linear r-permutations divided by r.

Example: Ten people, including two who do not wish to sit next to one another, are to be seated at a round table. How many circular seating arrangements are there?

4.3 Combinations of sets

4.3.1 Combinations of sets Definition: Let r be a non-negative integer.

An r-combination of a set S is an r-element subset of S. We denote by nCr, or C(n,r), or

r

n

Example: Let S={a,b,c,d}, then {a,b,c}, {a,b,d}, {a,c,d}, {b,c,d} are the 3-combinations of S.

Note that these are subsets, not sequences. Therefore,

{a,b,c}={a,c,b}={b,a,c}={b,c,a}={c,a,b} ={c,b,a}.

If r>n, then C(n,r)=0. Also C(0,r)=0 if r>0. Obviously

C(n,0)=1, C(n,1)=n, C(n,n)=1.

Theorem 4.7: For 0rn, C(n,r)=p(n,r)/r! and hence C(n,r)=n!/r!(n-r)!.

Proof: Let S be an n-element set. Each r-permutation of S arises in exactly one way as

a result of carrying out the following two tasks. (1)Choose r elements from S. C(n,r) (2)Arrange the chosen r elements in some order. r! By the multiplication principle we have p(n,r)=r!

C(n,r). Thus C(n,r)=p(n,r)/r!. We now use the formula p(n,r)=n!/(n-r)! and obtain C(n,r)=n!/r!(n-r)!.

Corollary 4.1: For 0rn, C(n,r)=C(n,n-r). Proof: C(n,r)=n!/(r!(n-r)!)=n!/((n-(n-r))!(n-

r)!)=C(n,n-r). Example: How many different seven-person

committees can be formed each containing three women from an available set of 20 women and four men from an available set of 30 men?

Task1: Choose three women from the set of 20 women. C(20,3)

Task2: Choose four men from an the set of 30 men. C(30,4)

By the multiplication principle,there are C(20,3)C(30,4).

4.3.2 The Binomial Coefficients and Identities The number C(n,r) have many important and

fascinating properties. C(n,r) is also called a binomial coefficient

because these numbers occur as coefficients in the expansion of powers of binomial expressions such as (a+b)n.

Theorem 4.8(Binomial theorem): Let x and y be variables, and let n be a nonnegative integer. Then

n

k

knkn yxknCyx0

),()(

Corollary 4.2: Let n be a nonnegative integer. Then

C(n,0)+C(n,1)+…+C(n,n)=2n

Proof ： Let x=y=1, by the Binomial theorem it follows that

Corollary 4.3:Let n be a positive integer. Then

C(n,0)-C(n,1)+C(n,2)-…+ (-1)nC(n,n)=0 Proof ： Let x=-1, and y=1, by the Binomial

theorem it follows that C(n,0)-C(n,1)+C(n,2)-…+ (-1)nC(n,n)=0

Remark: Corollary 4.3 implies that C(n,0)+C(n,2)+ …=C(n,1)+C(n,3)+ …

Theorem 4.9: Let m,n,r, and k be nonnegative integer. Then

)1,1(),()1( knCk

nknC

(2)C(n,k)=C(n-1,k)+C(n-1,k-1)(杨辉公式， Pascal’s formula)；

1

1

2),()3(

n

n

k

nknkC 2

1

2 2)1(),()4(

n

n

k

nnknCk

(5)C(n,r)C(r,k)=C(n,k)C(n-k,r-k), where rk；(6)C(m,0)C(n,r)+C(m,1)C(n,r-1)++C(m,r)C(n,0)=C(m+n,r)，where rmin{m,n} (Vandermonde identity)；(7)C(m,0)C(n,0)+C(m,1)C(n,1)++C(m,m)C(n,m)=C(m+n,m)where mn 。 When m=n，

),2(),(0

2 nnCknCn

k

),1(),()8(0

mmnCkknCm

k

4.4 Permutations and Combinations of multisets

Multisets :A multiset is a set in which an item may appear more than once. .

Sets and multisets are quite similar: both support the basic set operations of union, intersection, and difference.

item ai ni， {n1•a1,n2•a2,…,nk•ak} Example ： {a,a,a,a,b,b,c} {4•a,2•b,1•c} {•a1,•a2,…,•ak}

4.4.1 Permutations of multisets If S is a multiset, a r-permutation of S is an

ordered arrangement of r of the objects of S. If the total number of objects of S is n ， then a n-permutation of S will also be called a permutation of S.

For example, if S={2•a,1•b,3•c}, then aacb acbc cacc are 4-permutations of S, “abccac” is a permutation of S. The multiset S has no 7-permutations since

7>2+1+3=6, the number of objects of S.

Theorem 4.10: Let S be a multiset with objects of k different types where each has an infinite repetition number. Then the number of r-permutations of S is kr.

Proof: In constructing a r-permutation of S, we can choose the first item can be an object of any

one of the k types. Similarly the second item to be an object of any one

of the k types, and so on. Since all repetition numbers of S are infinite, the

number of different choices for any item is always k and does not depend on the choices of any previous items.

By the multiplication principle, the r items can be chose in kr ways.

Corollary 4.4: Let S={n1•a1,n2•a2,…,nk•ak} ， and ni r for each i=1,2,…,n ， then the number of r-permutations of S is kr.

Theorem 4.11: Let multiset S={n1•a1,n2•a2,…,nk•ak}, and n=n1+n2+…+nk=|S|. Then the number of permutations of S equals n!/(n1!n2!…nk!)。

Proof: We can think of it this way. There are n places, and we want to put exactly one of the objects of S in each of the places.

Since there are n1 a1’s in S, we must choose a subset of n1 places from the set of n places.

C(n,n1) We next decided which places are to be

occupied by the a2’.

Example What is the number of permutations of the letters in the word Mississippi?

Let S={n1•a1,n2•a2,…,nk•ak}, and n=n1+n2+…+nk=|S| ， then the number N of r-permutations of S equals

(1)0 r>n (2)n!/(n1!n2!…nk!) r=n (3)kr ni r for each i=1,2,…,n (4)If r<n, there is, in general, no simple

formula for the number of r-permutations of S.

Nonetheless a solution can be obtained by the technique of generating functions, and we discuss this in 4.6 .

Combinations of multisets P85-86 3.2 Inclusion-Exclusion principle and

Applications

Exercise P83 17, 19,22,33; P86 4,5,6,8 1.In how many ways can six men and six ladies be

seated round table if the men and ladies are to sit in alternate seats?

2.In how many ways can 15 people be seated at a round table if B refuse to sit next to A?What if B only refuses to sit on A’s right?

3.Prove Theorem 4.9 (4)(7). 4. How many ways are there to assign three jobs to

five employees if each employee can be given more than one job?

5.A book publisher has 3000 copies of a discrete mathematics book . How many ways are there to store these books in there three warehouses if the copies of the book are indistinguishable?