From Chapter 1 of Discrete and Combinatorial Mathematics, 4th ed, by R' P' Grimaldi

1
Fundamental Principles of Counting

• From Chapter 1 of Discrete and Combinatorial
  Mathematics, 4th ed, by R. P. Grimaldi

2
Introduction

• Enumeration does not end up with arithmetic.
• It also has applications in such areas as coding
  theory, probability and statistics, and in the
  analysis of algorithms.
• Be sure to learn and understand the basic
  formulas -- but do not rely on them to heavily.
• For without an analysis of each problem, a mere
  knowledge of formulas is next to useless.

3
1.1 The Rules of Sum and Product

• We want to develop the ability to decompose
  such problems and piece together our partial
  solutions in order to arrive at the final answer.
• The Rule of Sum If a first task can be be
  performed in m ways, while a second task can be
  performed in n ways, and the two task can be
  performed simultaneously, then performing either
  task can be accomplished in any one of mn ways.

4
1.1 The Rules of Sum and Product

• Example 1.1 40 textbooks on sociology and 50
  textbooks on anthropology
• Example 1.2 Five introductory books each on C,
  FORTRAN, Java, and Pascal
• Example 1.3 Two colleagues, one has three
  textbook on the analysis of algorithms, and the
  other has five. , n denotes the
  maximum number of different books on this topic
  that an instructor can borrow.

5
1.1 The Rules of Sum and Product

• Example 1.4

12 employees
Committee A 5
Committee B 7
Should the administrator decide to speak to just
one committee member before making her decision,
then 12 employee she can call upon for input.
A member of Committee A on Monday, a member of
Committee B on Tuesday, then .
6
1.1 The Rules of Sum and Product

• The Rule of Product
• If a procedure can be broken down into first and
  second stages, and if there are m possible
  outcomes for the first stage and if, for each of
  these outcomes, there are n possible outcomes for
  the second stage, then the total procedure can be
  carried out, in the designated order, in mn, ways.

7
1.1 The Rules of Sum and Product

• Example 1.6 License plate consisting of two
  letters followed by four digits
• No letter or digit cab be repeated
  26?25?10?9?8?7 32760
• With repetitions of letters and digits allowed
  26?26?10?10 ?10?106760000
• Example 1.7
• 1-byte address 2?2?2?2?2?2?2?2 256
• 2-byte address 256?25665536 available addresses
• 32-bite architecture 28?28?28?284,294,967,296
• 8-byte

8
1.1 The Rules of Sum and Product

• Example 1.8 Six kinds of muffins, eight kinds of
  sandwiches, and five beverages (hot coffee, hot
  tea,iced tea, cola, and orange juice). Want a
  lunch either a muffin and a hot beverage or a
  sandwich and a cold beverage.
• There are 6?2? 8?336 ways

9
1.2 Permutations

• Permutations
• Counting linear arrangements of objects
• We shall develop some systematic methods for
  dealing with linear arrangements.

10
1.2 Permutations

• Example 1.9 In a class of 10 students, five are
  to be chosen and seated in a row for a picture.
  How many such linear arrangements are possible?

10
9
8
7
6
?
?
?
?
30,240
11
1.2 Permutations

• Given a collection of n distinct objects, any
  (linear) arrangements of these objects is called
  a permutation of the collection.
• In general, if there are n distinct objects, the
  number of permutations of size r for the n
  objects is

We denote this number by P(n,r).
12
1.2 Permutations

• Example 1.11 The number of arrangements of four
  letters, BALL, is 12, not 4!.
• 2?(Number of arrangements of the letters, B, A,
  L, L)
• (Number of arrangements of the
  letters, B, A, L1, L2 )
• Example 1.12 The number of arrangements of four
  letters, PEPPER
• (2!)?(3!)(Number of arrangements of the letters
  in PEPPER)
• (Number of permutations of the symbols
  P1,E1,P2,P3,E2 ,R)
• ? (Number of arrangements of the letters in
  PEPPER)
• 6!/(2!)(3!)60

13
1.2 Permutations

• In general, if there are n objects with n1 of a
  first type, n2 of a second type, , nr of an rth
  type, where n1 ? n2 ?? nr n, then there are n!/
  n1!n2 ! nr! Arrangements of the n given objects.

14
1.2 Permutations

• Example 1.14

• How many paths from (2,1) to (7,4)?
• Each path corresponds with a list of five Rs
  and three Us.
• 8!/(5!)(3!)56.

5
4
3
2
1
3
4
5
6
7
1
2
15
1.2 Permutations

• Example 1.16 If six people, designated A, B, ,
  F, are seated around table, how many different
  circular arrangement are possible, if
  arrangements are considered the same when one can
  be obtained from the other by rotation?

• 6?(Number of circular arrangements of A, B, ,
  F) (Number of linear arrangements of A, B, ,
  F) 6!
• ?6!/6120

A
C
D
B
D
F
E
A
C
E
F
B
16
1.2 Permutations

• Example 1.17 Suppose now that the six people of
  Example 1.16 are three married couples and that
  A, B, and C are the females. We want to arrange
  the six people around the table so that the sexes
  alternate.

17
1.3 Combinations The Binomial Theorem

• Draw three cards from a standard deck, in
  succession without replacement, then by the rule
  of product, there are

possibilities.
18
1.3 Combinations The Binomial Theorem

• Each selection, or combination, of three cards,
  with no reference to order, corresponds to 3!
  permutations of three cards.

(3!)?(Number of selection of size 3 from a deck
of 52) Number of permutation of size 3 from
the 52 cards
19
1.3 Combinations The Binomial Theorem

• In general, if we start with n objects, each
  selection, or combination, of r of these objects,
  with no reference to order, corresponds to r!
  permutations of size r from the n objects.
• Thus the number of combinations of size r from a
  collection of size n, denoted C(n,r), where 0?r
  ?n, satisfies (r!)?C(n,r)P(n,r) and

20
1.3 Combinations The Binomial Theorem

• When dealing with any counting problem, we should
  ask ourselves about the importance of order in
  the problem.
• When order is relevant, we think in terms of
  permutations, and arrangements and the rule of
  product.
• When order is not relevant, combination should
  play a key role in solving the problem,

21
1.3 Combinations The Binomial Theorem

• Example 1.18 Invite 11 of the 20 committee
  member. Order is not important, so she can invite
  the the lucky 11 in C(20,11)20!/(11!9!) ways.
• Example 1.19
• Answer any 7 of 10 questions
• Answer 3 from the first 5 and 4 from the last 5.
• Answer 7 of the 10 question, where at least 3 are
  selected from the first 5.
• Answer 3 of the first 5 and 4 of the last 5.
• Answer 4 of the first 5 and 3 of the last 5.
• Answer all 5 of the first 5 and 2 of the last 5.

22
1.3 Combinations The Binomial Theorem

• Example 1.20
• Select 9 from 28 juniors and 25 seniors.
• If two juniors and one senior are the best
  spikers and must be on the team, then the rest of
  the team can be chosen in
• For a tournament the team must comprise 4 juniors
  and 5 seniors. She can select her team in

23
1.3 Combinations The Binomial Theorem

• Example 1.21 Make up four teams, termed A, B, C
  and D, of nine girls each from the 36 freshman
  girls.

• Team A
• Team B
• Team C
• Team D

• Distribute 9 As, 9 Bs, 9Cs, and 9 Ds in the
  36 spaces.
• The number of way is the number of arrangements
  of 36 letters comprising 9 each of A, B, C and D.

24
1.3 Combinations The Binomial Theorem

• Example 1.22
• The number of arrangements of letters in
  TALLAHASSEE is 11!/3!2!2!1!1!.
• How many of these arrangements have no adjacent
  As?
• Disregarding the As, there are
  8!/2!2!2!1!1!5040 ways
• Insert 3 As in 9 possible locations
• Each of the insertion is for all other 5039
  arrangements of other letters. By the rule of
  product, there are 5040?84423,360 ways.

25
1.3 Combinations The Binomial Theorem

• Example 1.23
• String made up of prescribed alphabets
• E.g. 01, 11, 21, 12, 22, 000, 012,202, 110 of 0,1
  and 2
• In general, if n is any positive integer, then
  by the rule of product there are 3n strings of
  length n for alphabet 0, 1, and 2.
• Define weight of x, wt(x), by wt(X)x1x2x3xn.
  
• Among the 310 strings of length 10, we wish to
  determine how many even weight.
• Such a string has even weight precisely when the
  number of 1s in the string is even.

Continued
26
1.3 Combinations The Binomial Theorem

• Example 1.23
• Six different cases to consider.
• If the string contains no 1s, then each of the
  10 positions can be filled with either 0 or 2
  210.
• Two 1s 28.
• Four 1s 26.
• Six 1s 24.

• Eight 1s 22.
• Ten 1s .
• By the rule of sum,

27
1.3 Combinations The Binomial Theorem

• Example 1.24
• Draw five cards from a standard deck of 52 cards.
  In how many ways can her selection result in a
  hand with no clubs?
• Ellen is restricted to selecting her five cards
  from the 39
• cards in the deck that are not clubs.

Continued
28
1.3 Combinations The Binomial Theorem

• We want to count the number of Ellens five card
  selection that contain at least one club.
• These are precisely the selections that were not
  counted in part (a).
• There are possible five-card hands in
  total, we find that

Continued
29
1.3 Combinations The Binomial Theorem

• Can we obtain the result in part (b) in another
  way?
• For example, since Ellen wants to have at least
  one club in the five-card hand, let her first
  selection a club.
• Now she does not care what comes up for other
  four cards.

• Something wrong!

3? 5? K? 7? J?
Continued
5? 3? K? 7? J?
Over-counting
K? 3? 5? 7? J?
30
1.3 Combinations The Binomial Theorem

• Any other way to arrive at the answer in part (b)?

31
1.3 Combinations The Binomial Theorem

• Theorem 1.1 The Binomial Theorem. If x and y are
  variables and n is a positive integer, then

Continued
32
1.3 Combinations The Binomial Theorem

• In n4, the coefficient of x2y2 in the expansion
  of the product (xy)(xy)(xy)(xy) is the number
  of ways in which we can select two xs from the
  four xs, one of which is available in each
  factor.
• For example, among the possibilities, we can
  select (1) x from the first two factors and y
  from the last two factors or (2) x from the first
  and third factors and y from the second and
  fourth.

• Factor selected for x
• 1,2
• 1,3
• 1,4
• 2,3
• 2,4
• 3,4

• Factor selected for y
• 3,4
• 2,4
• 2,3
• 1,4
• 1,3
• 1,2

Continued
33
1.3 Combinations The Binomial Theorem

• Consequently, the coefficient of x2y2 in the
  expansion of (xy)4 is 6, the number of ways
  to select two distinct objects from a collection
  of four distinct objects.

Continued
34
1.3 Combinations The Binomial Theorem

• Proof In the expansion of the product

The coefficient of xkyn-k, where 0?k?n, is the
number of different ways in which we can select k
xs from the n available factors. The total
number of such selections of size k from a
collection of size n is C(n,k) , and from
this the binomial theorem follows.
35
1.3 Combinations The Binomial Theorem

• Example 1.25
• From the binomial theorem it follows that the
  x5y2 coefficient of (xy)7 in the expansion of is
• To obtain the coefficient of a5b2 in the
  expansion of (2a-3b)7, replace 2a by x and 3b by
  y.

36
1.3 Combinations The Binomial Theorem

• Corollary 1.1 For each integer ngt0,

37
1.3 Combinations The Binomial Theorem

• Theorem 1.2 For positive integers n, t, the
  coefficient of in the
  expansion of
• is
• where each ni is an integer with 0? ni ? n, for
  all 1 ? i ? t, and
  .

38
1.3 Combinations The Binomial Theorem

• Example 1.26
• (xyz)7
• x2y2z3
• xyz5
• We need to know the coefficient of a2b3c2d5 in
  the expansion of (a3b-3c2d5)16.

39
1.4 Combinations with Repetition

• When repetitions are allowed, we have seen that
  for n distinct objects an arrangement of size r
  of these objects can be obtained in nr ways, for
  an integer r?0.
• We now turn to the comparable problem for
  combinations and once again obtain a related
  problem whose solution follows from out previous
  enumeration principles.

40
1.4 Combinations with Repetition

• Example 1.27 Seven high school freshmen, each
  wants one of the following a cheeseburger (c), a
  hot god (h), a taco (t), or a fish sandwich (f).
  How many different purchases are possible?
• Here we are concerned with how many of each item
  are purchased, not with the order in which they
  are purchased, so the problem is one of
  selections, or combinations, with repetition.

Enumerating all arrangements of 10 symbols
consisting of 7 xs and 3 s
41
1.4 Combinations with Repetition

• In general, when we wish to select, with
  repetition, r of n distinct objects, we find that
  we are considering all arrangements of r xs and
  n-1 s and that this number is

Consequently, the number of combination of n
objects taken r at a time, with repetition, is
C(nr-1,r).
42
1.4 Combinations with Repetition

• Example 1.28 A donut shop offer 20 kinds of
  donuts. Assuming that there are at least a dozen
  of each kind when we enter the shop, we can
  select a dozen in C(2012-1,12)C(31,12)141,120,5
  25 ways.

43
1.4 Combinations with Repetition

• Example 1.29 President wishes to distribute
  among four vice presidents 1000 in Christmas
  bonus checks, where each check will be written
  for a multiple of 100.
• One or more presidents get nothing.
  C(410-1,10)C(13,10)
• Each president should receive at least 100.
  C(46-1,6)C(9,6)
• If each vice president must get at least 100 and
  Mona gets at least 500, then
• C(32-1,2)C(31-1,1)C(30-1,0)10C(42-1,1)

Mona gets exactly 600
Mona gets exactly 600
Mona gets exactly 500
44
1.4 Combinations with Repetition

• Example 1.30 In how many ways can we distribute
  7 apples and 6 oranges among 4 children so that
  each children receives at least one apple?
• Given each child an apple, we have C(43-1,3)20
  ways to distribute the other 3 apples.
• There are C(46-1,6)84 ways to distribute 6
  oranges.
• By the rule of product, 20?841680 ways

45
1.4 Combinations with Repetition

• Example 1.31 A message is made up of 12
  different symbols and is to be transmitted
  through a communication channel. In addition to
  12 symbols, the transmitter will also send a
  total of 45 spaces between the symbols, with at
  least three spaces between each pair of
  consecutive symbols. In how many ways can the
  transmitter send such a message?

12!
C(1112-1,12)
?
45-11?312 A selection of size 12 from a
collection of size 11
arrangement of 12 different symbols
46
1.4 Combinations with Repetition

• Example 1.32Determine all integer solutions to
  the equation
  where xi?0 ? 1?i ?4.
• A selection, with repetition, of size 7 from a
  collection of size 4, so there are C(47-1,7)
  solutions.
• At this point it is crucial that we recognize the
  equivalence of the following
• The number of integer solutions of the equation
• The number of selections, with repetition, of
  size r from a collection of size n.
• The number of ways r identical objects can be
  distributed among n containers.

47
1.4 Combinations with Repetition

• Example 1.33 In how many ways can one distribute
  10 (identical) white marbles among 6 distinct
  containers?
• C(610-1,10)3003

48
1.4 Combinations with Repetition

• Example 1.34How many solutions are there to the
  inequality
• One approach is to determine the number of such
  solutions to
  where k is an integer and
  0?k?9.
• Transform the problem by noting the
  correspondence between the the nonnegative
  integer solutions of the above equation and the
  integer solution of
• The number of solutions is the same as
• where yixi, for 1?i ?6, and y7x7-1.

C(79-1,9)
49
1.4 Combinations with Repetition

• Example 1.36
• Let us determine all the different ways in which
  we can write the number 4 as a sum of positive
  integers, where the order of summands is
  considered relevant.

dont care order
50
1.4 Combinations with Repetition

• Example 1.36
• We wish to count the number of compositions for
  the number 7.
• Consider the number of possible summands.
• One summand 1
• If there are two (positive) summands, we want to
  count the number of integer solutions for
  w1w27, where w1,w2 gt0.
• This is equal to the number integer solutions
  for x1x25, where x1,x2 ?0.

Continued
51
1.4 Combinations with Repetition

• In general, one finds that for each positive
  integer m, there are
• compositions.

52
1.4 Combinations with Repetition

• Example 1.37 How many time is the print
  statement executed?

for i1 to 20 do for j1 to i do for
k1 to j do print(ijk)

• For the print statement to be executed, it must
  satisfy the condition 1?k?j?i?20.
• In fact, any selection a, b, c(a?b?c) of size 3,
  with repetition allowed, from the list 1, 2, 3,
  , 20 results in one of the correct selection
  here ka, jb, ic.