Discrete Mathematics

1
Discrete Mathematics

• Chapter 7 Advanced Counting Techniques

???? ????? ???(Lingling Huang)
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Outline

• 7.1 Recurrence Relations
• 7.2 Solving Linear Recurrence Relations
• 7.4 Generating Functions
• 7.5 Inclusion-Exclusion
• 7.6 Applications of Inclusion-Exclusion

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7.1 Recurrence Relations(????)

• Example 1. Let an be a sequence that
  satisfies the recurrence relation anan-1-an-2
  for n2,3,, and suppose that a03,and a15.
• Here a03 and a15 are the initial conditions.
• By the recurrence relation,
• a2 a1-a0 2
• a3 a2-a1 -3
• a4 a3-a2 -5
• Q1 Applications ?
• Q2 Are there better ways for computing the terms
  of an?

4

• ?Modeling with Recurrence Relations
• We can use recurrence relations to model
  (describe) a
• wide variety of problems.

Example 3. Compound Interest (??) Suppose that a
person deposits(??) 10000 ina saving account at
a bank yielding 11 per year with interest
compounded annually. How much will be in the
account after 30 years ?
Sol Let Pn denote the amount in the account
after n years. PnPn-1
0.11?Pn-11.11 ? Pn-1, ? P301.11 ? P29(1.11)2
? P28(1.11)30 ? P0 228922.97
P010000
5

• Example 5. (The Tower of Hanoi)
• The rules of the puzzle allow disks to be moved
  one at a time from one peg to another as long as
  a disk is never placed on top of a smaller disk.
  Let Hn denote the number of moves needed to solve
  the Tower of Hanoi problem with n disks. Set up
  a recurrence relation for the sequence Hn.
• Sol Hn2Hn-11, ( n-1? disk ??peg
  1?peg 3, ? n ? disk ? peg 1?peg 2, n-1?
  disk ?? peg 3?peg 2)

?? n ?disk?? peg 1 ?? peg 2
H11
6

• ??? Hn2Hn-11, H11
• ?Hn2Hn-11
• 2(2Hn-21)1
• 22Hn-221
• 22(2Hn-31)21
• 23Hn-3(2221)
• 2n-1H1(2n-22n-31)
• 2n-12n-21
• 2n-1

7

• Example 6. Find a recurrence relation and give
  initial conditions for the number of bit strings
  of length n that do not have two consecutive 0s.
  How many such bit strings are there of length 5
  ?
• Sol

Let an be the number of bit strings of length n
that do not have two consecutive 0s.
? an an-1an-2, n ? 3 a12 (string
0,1) a23 (string 01,10,11)
1
an-2 ?
0
1
? a3a2a15, a48, a513
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• Example 7. (Codeword enumeration)
• A computer system considers a string of decimal
  digits a valid codeword if it contains an even
  number of 0 digits. Let an be the number of valid
  n-digit codewords. Find a recurrence relation
  for an.
• Sol

? an 9an-1 (10n-1-an-1) 8an-1
10n-1 , n?2 a1 9
19
an-1 ?
10n-1 - an-1 ?
0
9

• ?an??

Exercise 3,23,25,27,29,41
(41???n)
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7.2 Solving Recurrence Relations

• Def 1. A linear homogeneous recurrence relation
  of
• degree k (i.e., k terms) with constant
  coefficients
• is a recurrence relation of the form
• where ci?R and ck?0

an c1an-1c2an-2ckan-k
Example 1 and 2. fn fn-1 fn-2
an an-5 an an-1 an-22
an nan-1 Hn 2Hn-1 1
(True, deg2)
(True, deg5)
(False, not linear)
(False , not linear)
(False, not homogeneous)
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Solving Linear Homogeneous RecurrenceRelations
with Constant Coefficients

• Theorem 1.
• Let an c1an-1 c2an-2 be a recurrence relation
  
• with c1,c2?R.
• If r2 - c1r - c2 0 (??characteristic
  equation) has two distinct roots r1 and r2.
• Then the solution of an is an a1r1n a2r2n
  ,
• for n0,1,2,, where a1 , a2 are constants.
• (a1 , a2???
  a0, a1??)

12

• Example 3.
• Whats the solution of the recurrence relation
• an an-1 2an-2
• with a02 and a17 ?
• Sol
• The characteristic equation is r2 r - 20.
• Its two roots are r1 2 and r2 -1.
• Hence ana1?2n a2 ?(-1)n .
• ?a0 a1a2 2, a12a1-a27
• ?a1 3, a2 -1
• ? an 3?2n - (-1)n.

r1?r2?????,?????
??a2 a1 2a0 11 a2 3?22 - 1 11
13

• Example 4. Find an explicit formula for the
  Fibonacci numbers.
• Sol
• fn fn-1 fn-2 , n ? 2, f00 , f11.
• The characteristic equation is r2 - r -
  10.
• Its two roots are ,
  .
• So we have

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• Thm 2.
• Let an c1an-1c2an-2 be a recurrence relation
• with c1,c2?R.
• If r2 - c1r - c2 0 has only one root r0 .
• Then the solution of an is
• an a1 ? r0n a2 ? n ? r0n
• for n0,1,2,, where a1 and a2 are constants.

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• Example 5.
• Whats the solution of an 6an-1 - 9an-2 with
  a01 and a16 ?
• Sol
• The root of r2 - 6r 9 0 is r0 3.
• Hence an a1.3n a2.n.3n .
• ?a0 a1 1
• a1 3a1 3a2 6
• ? a1 1 and a2 1
• ? an 3n n.3n

??a2 6a1 - 9a0 27 a2 32 2 ? 32
27
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• Thm 3.
• Let an c1an-1 c2an-2 ckan-k be a
  recurrence relation with c1, c2, , ck ? R.
• If rk - c1rk-1 - c2rk-2 -- ck 0 has k
  distinct roots r1, r2,, rk.
• Then the solution of an is
• an a1r1n a2r2n akrkn, for n 0, 1, 2,
  
• where a1, a2,ak are constants.

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• Example 6 (k 3)
• Find the solution of an 6an-1 - 11an-2
  6an-3
• with initial conditions a02, a15 and a215 .
• Sol
• The roots of r3 - 6r2 11r 6 0 are
• r1 1, r2 2, and r3 3
• ?an a1 ? 1n a2 ?2n a3 ?3n
• ?a0 a1 a2 a3 2
• a1 a1 2a2 3a3 5
• a2 a1 4a2 9a3 15
• ?an 1 - 2n 2 ? 3n

a1 1, a2 -1, a3 2
??a3 6a2 - 11a1 6a0 47 a3 1 - 23
2 ? 33 47
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• Thm 4.
• Let an c1an-1 c2an-2 ckan-k be a
  recurrence relation with c1, c2, , ck ?R.
• If rk - c1rk-1 - c2rk-2 - - ck 0 has t
  distinct roots r1, r2, , rt with
  multiplicities m1, m2, , mt respectively,
  where mi ? 1,?i, and m1 m2 mt k,
• then

where ai,j are constants for 1 ? i ? t and 0 ? j
? mi-1.
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???? ???????root?1 (2??),
-2 (3??),
3 (???)
??????????? an (a1,1 a1,2 ?n) ? 1n (a2,1
a2,2 ?n a2,3 ?n2) ? (-2)n a3,1 ? 3n
(????a?????1????,???????)
an (a1 a2 ?n) ? 1n (a3 a4 ?n a5 ?n2) ?
(-2)n a6 ? 3n
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• Example 8. Find the solution to the recurrence
  relation an -3an-1 - 3an-2 - an-3 with initial
  conditionsa0 1, a1 -2 and a2 -1.
• Sol
• r3 3r2 3r 1 0 has a single root r0
  -1 of multiplicity three.
• ? an (a1a2na3n2) r0n (a1a2na3n2)(-1)n
• ? a0 a1 1
• a1 (a1a2a3) ? (-1) -2
• a2 a12a24a3 -1
• ?a1 1, a2 3, a3 -2
• ? an (13n-2n2) ? (-1)n

??a3 - 3a2 - 3a1- a0 8 a3
(13?3-2?32)?(-1)3 8
Exercise 3,13,15,19
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Linear Nonhomogeneous RecurrenceRelations with
Constant Coefficients
Example an 3an-1 2n
A recurrence relation of the form an
c1an-1 c2an-2 ckan-k F(n), where c1,
c2, , ck are real numbersand F(n) is a function
not identically zero dependingonly on n.
The recurrence relation an c1an-1 c2an-2
ckan-k is called the associated homogeneous
recurrence relation.
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Example 9 an an-1 2n, associated
h.r.r ? an an-1 an an-1 an-2 n21,
associated h.r.r ? an an-1 an-2 an
3an-1 n3n, associated h.r.r ? an 3an-1
an an-1 an-3 n!, associated h.r.r ? an
an-1 an-3
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• Theorem 5. If an (p) is a particular solution
  (??) of an c1an-1 c2an-2 ckan-k F(n),
• then every solution is of the form an (p)
  an (h), where an (h) is a solution of
• an c1an-1 c2an-2 ckan-k

• Proof. If an (p) and bn are both solutions of
  
• an c1an-1 c2an-2 ckan-k
  F(n),
• then an(p) c1an-1(p) c2an-2 (p)
  ckan-k(p) F(n), and bn c1bn-1 c2bn-2
  ckbn-k F(n).
• ? an(p) - bn c1(an-1 - bn-1) c2(an-2 -
  bn-2)
  ck(an-k - bn-k)
• ? an(p) - bn is a solution of an c1an-1
  c2an-2 ckan-k
• ? bn an(p) an (h)

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• Example 10. Find all solutions of the recurrence
  relation an 3an-1 2n. What is the solution
  with a13?
• Sol
• associated??? an 3an-1 ??
• Characteristic equation r 3 0 ? r 3
  ? an(h) a ?3n.
• particular solution
• ? F(n) 2n ? Let an(p) cnd, where c, d
  ?R.
• If an(p) cnd is a solution to an 3an-1
  2n,
• then cnd 3(c(n-1)d)2n 3cn - 3c 3d 2n
• ? 2cn-3c 2d 2n (2c2)n (2d-3c) 0
  (??n?????0)
• ?2c2 0, and 2d-3c 0 ? c -1, d
  -3/2
• ? an(p) -n - 3/2 ? an an(h) an(p)
  a ?3n-n - 3/2
• If a1 a ?3-1 - 3/2 3 ? a 11/6
• ? an (11/6) ?3n- n - 3/2

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• Example 11. Find all solutions of the recurrence
  relation an 5an-1 -6an-2 7n.
• Sol
• associated??? an 5an-1 -6an-2 ??
• Characteristic equation r2 5r 6 0
• ? r1 3, r2 2
• ? an(h) a1 ? 3n a2 ? 2n.
• particular solution
• ? F(n) 7n ? Let an(p) c?7n, where c ?R.
• If an(p) c?7n is a solution to an 5an-1
  -6an-2 7n,
• then c?7n 5c?7n-1 - 6c?7n-2 7n
• ? 49c 35c - 6c 49
• ? c 49/20 ? an(p) (49/20) ?7n
• ? an an(h) an(p) a1 ? 3n a2 ? 2n
  (49/20) ?7n

Exercise 23
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• Theorem 6.
• an c1an-1 c2an-2 ckan-k F(n),
• where F(n) (btnt bt-1nt-1 b1n b0)sn.
• When s is not a root of the characteristic
  equationof the associated linear homogeneous
  recurrence relation, there is a particular
  solution of the form
• (ptnt pt-1nt-1 p1n
  p0)sn.
• When s is a root of the characteristic
  equation and its multiplicity is m, there is a
  particular solution of the form
• nm(ptnt pt-1nt-1 p1n
  p0)sn.

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• Example 12. What form does a particular solution
  of the linear nonhomogeneous recurrence relation
  an 6an-1 - 9an-2 F(n) have when F(n) 3n,
  F(n) n3n, F(n) n22n , and F(n) (n21)3n.
• Sol

The associated linear homogeneous recurrence
relation is an 6an-1 - 9an-2. characteristic
equation r2 - 6r 9 0 ? r 3 (2??)
F(n) 3n, and 3 is a root ? an(p) p0n23n
F(n) n3n, and 3 is a root ? an(p) n2(p1np0)
3n F(n) n22n , and 2 is not a root ? an(p)
(p2n2p1np0)2n F(n) (n21)3n , and 3 is a
root ? an(p) n2 (p2n2p1np0) 3n
Exercise 27
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• Example 13. Find the solutions of the recurrence
  relation an an-1 n with a11.
• Sol

The associated linear homogeneous recurrence
relation is an an-1 .
characteristic eq. r - 1 0 ? r 1 ? an(h)
c(1)nc
F(n) n n(1)n, and 1 is a root ? an(p)
n(p1np0)1n p1n2p0n
?an(p)??an an-1 n ? p1n2p0n
p1(n-1)2p0(n-1)n ? (2p1-1)np0-p10
? p1 ½, p0 p1 ½ ? an(p)
(n2n)/2 ? an an(p) an(h) (n2n)/2c
Exercise 29
a11 ? c0 ? an an(p) an(h) (n2n)/2
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• ex 40 Solve the simultaneous recurrence
  relations
• an 3an-1 2bn-1
• bn an-1 2bn-1 with a0
  1 and b0 2.
• Sol

an -bn 2an-1
? bn an-2an-1
? an 3an-1 2bn-1 3an-1 2an-1- 4an-2
? r 1, 4
? an 5an-1 - 4an-2
? r2 - 5r 4 0
? an a1a24n
a0 a1a2 1
? a1 -1, a2 2
a1 a14a2 3a0 2b0 7
? an 2?4n-1
? bn an-2an-1 2?4n-1-4n2 4n1
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7.4 Generating Functions.

• Def 1. The generating function for the sequence
  a0, a1, a2, of real numbers is the infinite
  series
• G(x) a0 a1x anxn
• (???an ?finite,????infinite,????????0)

31

• Example 1. Find the generating functions for the
  sequences ak with
• (1) ak 3
• (2) ak k1
• (3) ak 2k

Sol
(1) G(x)
(2) G(x)
(3) G(x)
32

• Example 2. What is the generating function for
  the sequence 1,1,1,1,1,1 ?
• Sol

G(x) a0 a1x a2x2 a3x3
(expansion,???)
(closed form)
Exercise 2
33

• Example 3.
• Let m?Z and ,for k 0, 1, , m.
• What is the generating function for the
  sequence a0, a1,, am ?
• Sol
• G(x) a0 a1x a2x2 amxm
• (1x)m (by????????)

34
Useful Facts About Power Series

• Example 4.
• The function f (x) is the
  generating
• function of the sequence 1, 1, 1, , because
  
• 1 x x2 when
  x lt 1.

Example 5. The function f (x) is
the generating function of the sequence 1,
a, a2, , because
1 ax a2x2 when ax lt 1 for
a?0.
Exercise 5(a)(b), 11(a)
35

• Theorem 1.
• Let f(x) and g(x) .
• Then f(x) g(x) .
• f(x) g(x) (a0a1xa2x2 )(b0b1x b2x2)
  (a0b0)(a0b1a1b0) x(a0b2a1b1a2b0) x2
  

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• Example 6.
• Let f (x) . Use Example 4 to find the
  coefficients a0, a1, a2, in the expansion f
  (x) .

Sol
? ak k1
37

• Def 2.
• Let u?R and k?N. Then the extended
• binomial coefficient is defined by

Example 7. Find and Sol
38

• Example 8
• When the top parameter is a negative integer,
  the extended binomial coefficient can be
  expressed in terms of an ordinary binomial
  coefficient.

39

• Thm 2. (The Extended Binomial Theorem)
• Let x?R with xlt1 and let u?R, then

40

• Example 9.
• Find the generating functions for (1x)-n and
  (1-x)-n where n?Z
• Sol By the Extended Binomial Theorem,

(??)
By replacing x by x we have
Exercise 11(b)(d)
41
Counting Problems and Generating Functions
Generating functions can be used to count the
numberof combinations of various types.

• Example 10.
• Find the number of solutions of e1 e2 e3
  17, where e1, e2, e3 are integers with 2? e1 ? 5,
  3? e2 ? 6, and 4? e3 ? 7.

Sol The number of solutions with the indicated
constraints is the coefficient of x17 in the
expansionof
(x2 x3 x4 x5)(x3 x4 x5 x6)(x4 x5
x6 x7)
(????? e1, e2, e3 ? xe1 xe2 xe3 x17)
?(e1, e2, e3)(4, 6, 7), (5, 5, 7), (5, 6, 6) ?3?
42
Example 11. In how many different ways can
eight identical cookies be distributed among
three distinct children if each child receives at
least two cookies and no more than four cookies?
Sol The number of solutions is the coefficient
of x8 in the expansion of
(x2 x3 x4)3
?(c1, c2, c3) (2, 2, 4), (2, 3, 3), (2, 4, 2),
(3, 2, 3), (3, 3, 2), (4,
2, 2) ?6?
Exercise 23
43
?Using Generating Functions to solve
Recurrence Relations.

• Example 16.
• Solving the recurrence relation ak 3ak-1 for
  k1,2,3, and initial condition a0 2.
• Sol
• ?? (by 7.2?Thm 1??)
• r 3 0 ? r 3 ? an a ? 3n
• ? a0 2 a
• ? an 2 ? 3n

44

• Let
  be the
• generating function for ak.
• First note that ak 3ak-1
• ?
• ? G(x) - a0 3x ? G(x)
• ?a0 2 ? G(x) - 3x ? G(x) G(x)(1-3x) 2

? ak 2 ? 3k
Exercise 5,7,11,33
45
Example 17 Solving ak 8ak-1 10k-1 for k
1,2,3, and initial condition a1 9. (Sec. 7.1
Example 7)
Sol Let a0 1 (????????).
Let
be the generating function for ak.
46
? ak (10k 8k)/2
Exercise 33
47
7.5 Inclusion-Exclusion ????

• A,B,C,D sets

1
2
2
1
1
3
1
2
0
1
1
2
1
48

• Theorem 1.
• A1, A2, , An sets

Exercise 17
49
7.6 Applications of Inclusion and Exclusion

• Example 1. How many solutions does x1 x2 x3
  11have, where x1, x2, x3 are nonnegative
  integers withx1 ? 3, x2 ? 4, and x3 ? 6?
• Sol

Let a solution have property P1 if x1 ? 4,
property P2 if x2 ? 5, and property P3 if x3 ? 7.
N(P1P2P3) N - N(P1) - N(P2) - N(P3)
N(P1P2) N(P2P3)
N(P1P3) - N(P1P2P3)
50
The Number of Onto Functions

• Example 2. How many onto functions are there
  form set A1, 2, 3, 4, 5, 6 to set Ba, b, c
  ?
• Sol f A ? B

f (1) a, b, c f (2) ? ? f
(6)
???????????? ???a,b,c??? ?
The number of onto functions (??????) -
(a,b,c????????) (a,b,c???????) -
(a,b,c?????)
51

• Thm 1. A m , B n
• There are
• onto functions f A ? B.

pf A a1, a2, , am. B b1, b2, ,
bn f (a1) f (a2)
? ? f (am)
b1, b2, , bn
Exercise 8
52

• Example 3. How many ways are there to assign five
  different jobs to four different employees if
  every employee is assigned at least one job?
• Sol

Consider the assignment of jobs as a function
from the set of five jobs to the set of four
employees.
The number of onto functions
Exercise 9
53
Derangements (??)

• Def. A derangement is a permutation of objects
  that leaves no object in its original position.

Example 5. The permutation 21453 is a
derangement of 12345. But 21543 is not a
derangement of 12345.
54
Def. Let Dn be the number of derangements of n
objects.
D2 1 because the derangements of 12 are 21.
D3 2 because the derangements of 123 are 231
and 312.
D4 (??4????permutation?) -
(4???????????permutation?)
(4???????????permutation??) -
(4????????????permutation??)
(4????????permutation??)
55
Theorem 2. (????)
Proof.
Exercise 13
56
Exercise 17. How many ways can the digits 0,
1, 2, 3, 4, 5, 6, 7, 8, 9 be arranged so that
even digit is in its original position?
Sol