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4.2 Permutations of sets

4.2 Permutations of sets. 4.2.1 Basic counting principles Theorem 4.3(Addition principle): If A 1 , A 2 , … , A n are disjoint sets, then the number of elements in the union of these sets is the sum of the numbers of elements in them. | A 1 ∪A 2 ∪…∪A n |=|A 1 |+|A 2 |+…+|A n |.

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4.2 Permutations of sets

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  1. 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∪A2∪…∪An |=|A1|+|A2|+…+|An|

  2. 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.

  3. 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. 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.

  5. Theorem 4.5: For n and r positive integers with rn, • 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)!.

  6. 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! .

  7. 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

  8. linar permutation • circular permutation • linar permutation 12345 • linar permutation 45123 • circular permutation

  9. For example, the circular permutation • arises from each of the linear permutation • 12345 23451 34512 45123 51234

  10. 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.

  11. 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?

  12. 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

  13. 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.

  14. Theorem 4.7: For 0rn, 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)!.

  15. Corollary 4.1: For 0rn, 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).

  16. 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

  17. 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

  18. 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)+ …

  19. (2)C(n,k)=C(n-1,k)+C(n-1,k-1)(杨辉公式，Pascal’s formula)； • Theorem 4.9: Let m,n,r, and k be nonnegative integer. Then (5)C(n,r)C(r,k)=C(n,k)C(n-k,r-k), where rk； (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 rmin{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 mn。When m=n，

  20. 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}

  21. 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.

  22. 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.

  23. 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.

  24. 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’.

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

  26. 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 .

  27. Combinations of multisets P85-86 3.2 • Inclusion-Exclusion principle and Applications

  28. 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?

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