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4.4 Permutations and Combinations of multisets

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 a i n i ，

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4.4 Permutations and Combinations of multisets

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  1. 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}

  2. 4.4 Permutations and Combinations of multisets • 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.

  3. 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.

  4. 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.

  5. 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’.

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

  7. 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 .

  8. 4.4.2 Combinations of multisets • If S is a multiset, a r-combination of S is an unordered selection of r of the objects of S. • Thus an r-combination of S is a submultiset of S. If S has n objects，then there is only one n-combination of S, namely, S itself. • If S contains objects of k different types, then there are k 1-combinations of S. • Example If S={2•a,1•b,3•c}, then the 3-combinations of S are • {2•a,1•b},{2•a,1•c}, {1•a,1•b,1•c}, {1•a,2•c},{1•b,2•c},{3•c}.

  9. Theorem 4.12: Let S ={·a1,·a2,…, ·ak}. Then the number of r-combinations of S equals C(k+r-1,r)。 • Proof. (1)The number of r-combinations of S equals the number of solutions of the equation • where x1,x2,…,xk are non-negative integers (2)We show that the number of these solutions equals the number of permutations of the multiset T={(k-1)·0,r·1}.

  10. Corollary 4.5: Let S={n1•a1,n2•a2,…,nk•ak}，and nir for each i=1,2,…,n. Then the number of r-combinations of S is C(k+r-1,r). • Example Suppose that a cookie shop has seven different kinds of cookies. How many different ways can twelve cookies be chosen?

  11. Example How many integer solutions does the equation x1+x2+x3=11 have, where x1,x2,and x3 are non-negative integers? • Solution To count the number of solutions, we note that a solution corresponds to a way of selecting 11 items from a set with three elements.

  12. Example How many solutions are there to the equation x1+x2+x3=11 where x13,x21, and x30? • Solution We introduce new variables: • y1=x1-3, y2=x2-1, y3=x3, • and the equation becomes y1+y2+y3=7 where y1,y2,and y3 are non-negative integers

  13. Corollary 4.6 Let S={·a1,·a2,…,·ak},and rk . Then the number of r-combinations of S so that each of the k types of objects occurs at least once equals C(r-1,k-1). • Proof: r-combinations of S so that each of the k types of objects occurs at least once, a1,a2,…,ak， • r-k combinations • r-k combinations, +a1,a2,…,ak，r-combinations of S so that each of the k types of objects occurs at least once

  14. Let S={n1•a1,n2•a2,…,nk•ak}, and n=n1+n2+…+nk=|S|，then the number N of r-combinations of S equals • (1) 0 r>n • (2) 1 r=n • (3) N=C(k+r-1,r) nir for each i=1,2,…,n. • (4) If r<n, and there is, in general, no simple formula for the number of r-combinations of S. Nonetheless a solution can be obtained by the inclusion-exclusion principle and technique of generating functions, and we discuss these in 4.5 and 4.6.

  15. 4.5 Inclusion-Exclusion principle and Applications • 4.5.1 Inclusion-Exclusion principle • Theorem 4.13：Let A and B be finite sets. Then |A∪B|=|A|+|B|-|A∩B|。 • Proof：Because A∪B=A∪(B-A)，and A∩(B-A)=,by theorem we obtain |A∪B|=|A|+|B-A|. • |B-A|=？ • Theorem 4.14：Let A1,A2,…,An be finite sets. Then

  16. Corollary 4.7: Let S be a finite set, and P1,P2,…,Pn be n properties referring to the objects in S. Let • Ai={x|xS and x has property Pi}(i=1,2,…,n) be the subset of S which have property Pi (and possibly other properties). Then the number of objects of S which have none of the properties P1,P2,…,Pn is given by

  17. Example: In a certain school, the pupils have to study France, German, or English. Each pupil must study at least one of the three. Among a group of 100 pupils, 42 are studying France, 45 are studying German, 65 are studying English, 15 are studying France and German, 20 are studying France and English, 25 are studying German and English. Find the number of pupils who are studying all the three subjects. And Find the number of pupil who are studying English only. • A, France • B, German • C, English • the number of pupil who are studying English only • |C|-|A∩C|-|B∩C|+|A∩B∩C|=28

  18. Example：How many hex strings of length r contain 0,1, and 2? • Solution：Let S be the set of length r’s hex strings. • Let A be the set of length r’s hex strings which do not contain 0 • Let B be the set of length r’s hex strings which do not contain 1 • Let C be the set of length r’s hex strings which do not contain 2.

  19. Example：Find the number of integers between 1 to 1000, inclusive, which are divisible by none of 5,6, and 8. • Solution: Let S be the set consisting of the first thousand positive integers. • Let P1 be the property that an integer is divisible by 5. • Let P2 be the property that an integer is divisible by 6. • Let P3 be the property that an integer is divisible by 8. • For i=1,2,3 let Ai be the set consisting of those integers in S with property Pi.

  20. Integers in the set A1∩A2 are divisible by both 5 and 6(lcm{5,6}=30). Integers in the set A1∩A3 are divisible by both 5 and 8(lcm{5,8}=40). Integers in the set A2∩A3 are divisible by both 6 and 8(lcm{6,8}=24). • Integers in the set A1∩A2∩A3 are divisible by 5 , 6 and 8(lcm{5,6,8}=120). • By the inclusion-exclusion principle, the number of integers between 1 to 1000 that are divisible by none of 5,6, and 8 equals

  21. Next: Applications of Inclusion-Exclusion principle, Generating functions, • Exercise P96 22,23 (Sixth) OR P83 22,33(Fifth); • P99 10,12,14,16, 18,20(Sixth) OR P86 10,12,14,16,18, 20(Fifth) • 1.Prove Theorem 4.9 (4)(7). • 2.How many solutions are there to the equation x1+x2+x3+x4+x5=25,where x1,x2,x3,x4, and x5 are nonnegative integers? • 3. How many strings of 20 decimal digits are there that contain two 0’s,four 1’s,three 2’s,one 3,two 4’s, three 5’s,two 7’s, and three 9’s? • 4.How many positive integers less than 1,000,000 have the sum of their digits equal to 19? • 5.A football team of 11players is to be selected from a set of 16 players, 5 of whom can only play in the backfield, 9 of whom can only play on the line, and 2 of whom can play either in the backfield or on the line. Assuming a football team has 7 men on the line and 4 in the backfield, determine the number of football teams possible. • 6. Find the number of integers between 1 and 10,000 inclusive which are not divisible by 4,5, or 6.

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