Transcript 离散数学7.ppt

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.

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



(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

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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
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|
 Theorem
 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 rpermutations 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 nelement 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
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
Theorem 4.6: The number of circular rpermutations 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 rpermutations, 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 relement subset of S. We denote by nCr, or
C(n,r), or  n 
r
 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))!(nr)!)=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
n
integer. Then
( x  y ) n   C ( n, k ) x k y n  k
k 0
 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)C (n, k ) 
n
C (n  1, k  1)
k
(2)C(n,k)=C(n-1,k)+C(n-1,k-1)(杨辉公式，Pascal’s formula)；
n
(3)  kC (n, k ) 
n 2n 1
k 1
n
(4)  k 2C (n, k )  n(n  1)2n  2
k 1
(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
n
mn。When m=n，
2
 C ( n, k )
k 0
m
(8) C (n  k , k )  C (n  m  1, m)
k 0
 C ( 2 n, n )
 Exercise
P96 17,19; P99 4,5,6,8 (Sixth)
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?
 OR
Next: Permutations and Combinations of
multisets