Pigeonhole Principle: Strong Form

4.1.2 Pigeonhole principle:Strong Form • Theorem 4.2: Let q1,q2,…,qn be positive integers. If q1+q2+…+qn-n+1 objects are put into n boxes, then either the first box contains at least q1 objects, or the second box contains at least q2 objects, … , or the nth box contains at least qn objects. • Proof:Suppose that we distribute q1+q2+…+qn-n+1 objects among n boxes.

(1)If n(r-1)+1 objects are put into n boxes, then at least one of the boxes contains r or more of the objects. Equivalently, • (2)If the average of n non-negative integers m1,m2,…,mn is greater than r-1: (m1+m2+…+mn)/n>r-1, then at least one of the integers is greater than or equal to r. • Proof:(1)q1=q2=…=qn=r • q1+q2+…+qn-n+1=rn-n+1=(r-1)n+1, then at least one of the boxes contains r or more of the objects。 • (2)(m1+m2+…+mn)>(r-1)n， • (m1+m2+…+mn)≥(r-1)n +1

Example 6:Two disks, one smaller than the other, are each divided into 200 congruent sectors. In the larger disk 100 of the sectors are chosen arbitrarily and painted red; the other 100 of the sectors are painted blue. In the smaller disk each sector is painted either red or blue with no stipulation on the number of red and blue sectors. The small disk is then placed on the larger disk so that their centers coincide. Show that it is possible to align the two disks so that the number of sectors of the small disk whose color matches the corresponding sector of the large disk is at least 100. • if the large disk is fixed in place • there are 200 possible positions for the small disk such that each sector of the small disk is contained in a sector of the large disk. • color matches the corresponding • 20000/200=100>100-1 • Position with at least 100 color matches

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|

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 a 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 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)!.

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

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 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)!.

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).

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

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)+ …

(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，

Exercise P96 17,19; P99 4,5,6,8 (Sixth) • OR P83 17, 19; P86 4,5,6,8(Fifth) • 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? • Next: Permutations and Combinations of multisets

Pigeonhole Principle: Strong Form