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}.

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

rxk

ii

1

(2)We show that the number of these solutions equals the number of permutations of the multiset T={(k-1)·0,r·1}.

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?

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.

Example How many solutions are there to the equation x1+x2+x3=11 where x13,x21, and x30?

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

Corollary 4.6 Let S={·a1,·a2,…,·ak},and rk . 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

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.

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||)1(|||||| 21

1

11n

n

jiji

n

ii

n

ii AAAAAAA

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|xS 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

||)1(

||||||||

21

121

nn

jiji

n

iin

AAA

AAASAAA

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

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.

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.

321 AAA

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

6008)412533()125166200(1000||

|)||||(||)||||(|||||

321

323121321321

AAA

AAAAAAAAASAAA

4.5.2 Applications of Inclusion-Exclusion principle

The number of r-combinations of multiset S 1. The number of r-combinations of multiset

S 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 4.5 .

Example: Determine the number of 10-combinations of multiset S={3·a,4·b,5·c}.

Solution ： We shall apply the inclusion-exclusion principle to the set Y of all 10-combinations of the multiset D={·a, ·b, ·c}.

Let P1 be the property that a 10-combination of D has more than 3 a’s. Let P2 be the property that a 10-combination of D has mote than 4 b’s. Let P3 be the property that a 10-combination of D has mote than 5 c’s.

For i=1,2,3 let Ai be the set consisting of those 10-combinations of D which have property Pi.

The number of 10-combinations of S is then the number of 10-combinations of D which have none of the properties P1, P2, and P3.

321 AAA

The set A1 consists of all 10-combinations of D in which a occurs at least 4 time.

If we take any one of these 10-combinations in A1 and remove 4 a’s, we are left with a 6-combination of D.

Conversely, if we take a 6-combination of D and add 4 a’s to it, we get a 10-combination of D in which a occurs at least 4 times.

Thus the number of 10-combinations in A1 equals the number of 6-combinations of D. Hence, |A1|=C(3+6-1,6)=C(8,6)=C(8,2) ，

Example: What is the number of integeal solutions of the equation

x1+x2+x3=5 which satisfy 0x12,0x22,1x35? Solution: We introduce new variables, x3'=x3-1 and our equation becomes x1+x2+x3'=4. The inequalities on the xi and x3' are satisfied if and

only if 0x12,0x22, 0x3'4. 4-combinations of multiset {2·a,2·b,4·c}

2.Derangements A derangement of {1,2,…,n} is a permutation

i1i2…in of {1,2,…,n} in which no integer is in its natural position:

i11,i22,…,inn.

We denote by Dn the number of derangements of {1,2,…,n}.

Theorem 4.15 ： For n1,)

!

1)1(

!3

1

!2

1

!1

11(!

nnD n

n

Proof: Let S={1,2,…,n} and X be the set of all permutations of S. Then |X|=n!.

For j=1,2,…,n, let pj be the property that in a permutation, j is in its natural position. Thus the permutation i1,i2,…,in of S has property pj

provided ij=j. A permutation of S is a derangement if and only if it has none of the properties p1,p2,…,pn.

Let Aj denote the set of permutations of S with property pj ( j=1,2,…,n).

Example:(1)Determine the number of permutations of {1,2,3,4,5,6,7,8,9} in which no odd integer is in its natural position and all even integers are in their natural position.

(2) Determine the number of permutations of {1,2,3,4,5,6,7,8,9} in which four integers are in their natural position.

3. Permutations with relative forbidden position

A Permutations of {1,2,…,n} with relative forbidden position is a permutation in which none of the patterns i,i+1(i=1,2,…,n) occurs. We denote by Qn the number of the permutations of {1,2,…,n} with relative forbidden position.

Theorem 4.16 ： For n1, Qn=n!-C(n-1,1)(n-1)!+C(n-1,2)(n-2)!-…+(-1)n-1

C(n-1,n-1)1!

Proof: Let S={1,2,…,n} and X be the set of all permutations of S. Then |X|=n!.

j(j+1), pj

Aj: pj

Qn=Dn+Dn-1

Generating functions

Exercise P86 6,8,10,12,14,16,18,20 1.How many solutions are there to the equation x1+x2+x3+x4+x5=25,where

x1,x2,x3,x4, and x5 are nonnegative integers? 2.How many strings of 20 decimal digits are there that contain two 0s,four

1s,three2s,one 3,two4s, three 5s,two 7s, and three9s? 3.How many positive integers less than 1,000,000 have the sum of their digits

equal to 19? 4.A football team of 11players is to be selected from a set of 15 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.

5. Find the number of integers between 1 and 10,000 inclusive which are not divisible by 4,5, or 6.

6. Determine the number of 12-combinations of the multiset S={4·a,3·b,5·c, 4·d }.

7. Determine the number of solutions of the equation x1+x2+x3+x4=14 in nonnegative integers x1,x2,x3, and x4 not exceeding 8.

8. Determine the number of permutations of {1,2,3,4,5,6,7,8} in which no even integer is in its natural position.

9. Determine the number of permutations of {1,2,…,n} in which exactly k integers are in their natural positions.

10.Eight boys are seated around a carousel. In how many ways can they change seats so that each has a different boy in front of him?