Lecture 4 part 1

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Lecture 4 (part 1)

• Combinatorics
• Reading Epp Chp 6

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Outline

• Basic Rules
• Linear Series Rule
• Multiplication Rule
• Addition and Difference Rule
• Inclusion-Exclusion Rule
• Common Counting Scenarios
• Permutations (Ordered Selections)
• Combinations (Unordered Selections)
• Counting ordered partitions / permutations of
  multi-sets
• Algebra of Combinations
• Binomial Theorem

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1. Basic Rules

• 1.1 Linear Series Rule.
• If m and n are integers such that m n, then
  there are n-m1 integers from m to n inclusive.

Example 1 How many integers are there in the
sequence 10, 11, 12, , 19, 20? Ans 20 10 1
11 Example 2 How many integers are there in
the sequence -8,-7,-6,-1,0,1,2,4, 5? Ans 5
(8) 1 14
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1. Basic Rules

• 1.1 Linear Series Rule.
• If m and n are integers such that m n, then
  there are n-m1 integers from m to n inclusive.

Example 3 How many integers are there in 0 to
1000 inclusive, that are divisible by 3. The
integers are 0,3,6,9,,996,999 , which are in the
form 3k, where k 0,1,2,,333. Hence there are
333-01 334 integers from 0 to 1000 which are
divisible by 3.
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1. Basic Rules Multiplication Rule

• 1.2 Multiplication Rule
• If an operation consists of a sequence of
  steps/events E1, E2 Ek and if each Ei can be
  performed in ni ways regardless of how the
  previous steps E1, Ei-1 were performed (i.e.
  independent), then the entire operation can be
  performed in n1n2nk ways.

Proof of multiplication rule is by
induction (Left as an exercise)
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1. Basic Rules Multiplication Rule

• 1.2 Multiplication Rule
• If an operation consists of a sequence of
  steps/events E1, E2 Ek and if each Ei can be
  performed in ni ways regardless of how the
  previous steps E1, Ei-1 were performed (i.e.
  independent), then the entire operation can be
  performed in n1n2nk ways.

Example 1 At breakfast, you are given a choice
of coffee or tea, and a choice of a scrambled,
fried and boiled egg. How many different kinds
of breakfast can you have?
Ans
Step 1 Choose coffee or tea gt 2 ways
Step 2 Choose how you would like your egg to be
done gt 3 ways regardless of the choice made in
step 1.
Total number of ways 2 x 3 6 ways (Mul. Rule)
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1. Basic Rules Multiplication Rule

• Example 2
• Let A and B be finite sets. If A m and B
  n, how many elements are there in (i) A B ?
  (ii) P(A)?

(i) The task of selecting pairs for A B is
reduced to the following steps Step 1 Choose
an element x from A m ways. Step 2 Choose an
element y from B n ways regardless of choices
made in step 1. Total number of ways of
performing the task total number of elements
in A B m n (Multiplication Rule)
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1. Basic Rules Multiplication Rule
Example 2 Let A and B be finite sets. If A
m and B n, how many elements are there in (i)
A B ? (ii) P(A)?

• (ii) Let the elements of A be a1 , a2 , a3 ,, am
• The task of forming a subset of A is reduced to
  the following procedure
• Step 1 Either take or drop element a1 2 ways.
• Step 2 Either take or drop element a2 2 ways
  regardless of choices made in step 1.
• Step 3 Either take or drop element a3 2 ways
  regardless of choices made in steps 1 and 2.
• Step m Either take or drop element am 2 ways
  regardless of choices made in steps 1, 2, 3, 4,
  , m-1.
• Total number of ways of performing the task (by
  multiplication rule) 2 2 2 2

2m
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1. Basic Rules Multiplication Rule

• Example 3
• A car license plate has 3 letters of alphabet
  followed by 4 single digit numbers. How many
  different car licenses can be issued (a) if
  repetitions of alphabets/numbers in the license
  plate is allowed (b) if repetitions are not
  allowed?

Ans (a) The task of forming a license plate
number consists of the following sub-tasks Step
1 choose the 1st letter 26 ways. Step 2
choose the 2nd letter 26 ways (independent of
step 1). Step 3 choose the 3rd letter 26 ways
(independent of steps 1-2). Step 4 choose the
1st digit 10 ways (independent of steps
1-3). Step 5 choose the 2nd digit 10 ways
(independent of steps 1-4). Step 6 choose the
3rd digit 10 ways (independent of steps
1-5). Step 7 choose the 4th digit 10 ways
(independent of steps 1-6). Total number of
performing the task 263 104
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1. Basic Rules Multiplication Rule

• Example 3
• A car license plate has 3 letters of alphabet
  followed by 4 single digit numbers. How many
  different car licenses can be issued (a) if
  repetitions of alphabets/numbers in the license
  plate is allowed (b) if repetitions are not
  allowed?

Ans (b) The task of forming a license plate
number consists of the following sub-tasks Step
1 choose the 1st letter 26 ways. Step 2
choose the 2nd letter 25 ways (independent of
step 1). Step 3 choose the 3rd letter 24 ways
(independent of steps 1-2). Step 4 choose the
1st digit 10 ways (independent of steps
1-3). Step 5 choose the 2nd digit 9 ways
(independent of steps 1-4). Step 6 choose the
3rd digit 8 ways (independent of steps
1-5). Step 7 choose the 4th digit 7 ways
(independent of steps 1-6). Total number of
performing the task 26 25 24 10 9 8
7
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1. Basic Rules Multiplication Rule

• Example 4 (Limitations of the Multiplication
  Rule)
• Two teams A and B are to play each other
  repeatedly until one wins two games in a row, or
  until a total of 3 games. How many ways can a
  tournament be played?

• Incorrect answer
• Step 1 Either A wins or B wins 2 ways.
• Step 2 Either A wins or B wins 2 ways.
• Step 3 Either A wins or B wins 2 ways.
• Total number of ways 2 2 2

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1. Basic Rules Multiplication Rule

• Example 4 (Limitations of the Multiplication
  Rule)
• Two teams A and B are to play each other
  repeatedly until one wins two games in a row, or
  until a total of 3 games. How many ways can a
  tournament be played?

• The above example illustrates a situation when
  the multiplication rule cannot be used.

• This is because the subsequent events (games) are
  dependent on the outcome of the previous events.

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1. Basic Rules Multiplication Rule

• Example 5 (Goal Re-ordering is a possible way to
  by-pass dependency of tasks)
• Three officers A president, a treasurer, and a
  secretary, are to be chosen from among 4 people
  A, B, C, D. Suppose that A cannot be president,
  and either C or D must be secretary. How many
  ways can the officers be chosen?

Incorrect answer Selecting three officers can
be broken down to the following tasks Step 1
Select the president 3 ways. Step 2 Select
the treasurer 3 ways. Step 3 Select the
secretary 2 ways. Total number of performing
the task 3 3 2 ways
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1. Basic Rules Multiplication Rule

• Example 5 (Goal Re-ordering is a possible way to
  by-pass dependency of tasks)
• Three officers A president, a treasurer, and a
  secretary, are to be chosen from among 4 people
  A, B, C, D. Suppose that A cannot be president,
  and either C or D must be secretary. How many
  ways can the officers be chosen?

Step 1 Select the president 3 ways
Step 2 Select the treasurer number of ways
DEPENDENT on the outcome of step 1!!!
Step 3 Select the secretary number of ways
DEPENDENT on the outcome of step 2, which is in
turn, DEPENDENT on step 1.
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1. Basic Rules Multiplication Rule

• Example 5 (Goal Re-ordering is a possible way to
  by-pass dependency of tasks)
• Three officers A president, a treasurer, and a
  secretary, are to be chosen from among 4 people
  A, B, C, D. Suppose that A cannot be president,
  and either C or D must be secretary. How many
  ways can the officers be chosen?

Answer Independence of events can sometimes be
restored through the re-ordering of your
tasks. Step 1 Select the secretary 2 ways (C
or D). Step 2 Select the president 2 ways
regardless of the choice taken in step 1 (person
B with the remaining person from step 1) Step 3
Select the treasurer 2 ways. (person A with the
person remaining from step 2) Total number of
performing the task 2 2 2 ways
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1. Basic Rules Multiplication Rule

• Example 5 (Goal Re-ordering is a possible way to
  by-pass dependency of tasks)
• Three officers A president, a treasurer, and a
  secretary, are to be chosen from among 4 people
  A, B, C, D. Suppose that A cannot be president,
  and either C or D must be secretary. How many
  ways can the officers be chosen?

Step 1 Select the secretary 2 ways
Step 2 Select the president 2 ways
Step 3 Select the treasurer 2 ways
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1. Basic Rules Multiplication Rule

• Summary
• Always check that the number of ways of each step
  is independent of the choices of the previous
  steps.
• The multiplication rule did NOT say
• Each step is independent of the choices of
  previous step.
• It says
• The number of ways of each step is independent
  of the choices of the previous step
• Get the subtle difference?
• Goal re-ordering may help.

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1. Basic Rules Addition/Difference Rule

• 1.3 Addition Rule
• If a finite set A A1 È A2 È...È An where all
  the Ais are mutually disjoint, then A A1
  A2 ... An
• Difference Rule
• If A is a finite set and BÍA, then A B A
  B

Addition Rule is proven by induction. (Proof are
left as exercise). Difference Rule is a corollary
of the addition rule. (corollary meaning that
it is a result of meaning that you can
prove/infer the difference rule from the addition
rule.) Because when BÍA, (A B) and B form a
partition of A (Meaning that (A B) È B A and
(A B) Ç B Æ) Therefore by addition rule A
B B A And so A B A B
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1. Basic Rules Addition/Difference Rule

• Example 1 A computer access code word consists
  of one to three letters, chosen from the 26
  alphabets, with repetitions allowed. How many
  code words are possible?

• Answer
• The set of all code words can be partitioned into
  subsets consisting of those of length 1, length
  2, and length 3.

• By addition rule Set of all code words of
  length 3 Set of all code words of
  length 1 Set of all code words of length
  2 Set of all code words of length 3

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1. Basic Rules Addition/Difference Rule

• Example 1 A computer access code word consists
  of one to three letters, chosen from the 26
  alphabets, with repetitions allowed. How many
  code words are possible?

• Answer
• The set of all code words can be partitioned into
  subsets consisting of those of length 1, length
  2, and length 3.

Set of all code words of length 3
Set of all code words of length 1
Set of all code words of length 2
Set of all code words of length 3

• Number of code words of length 1 26
• Number of code words of length 2 262
  (Multiplication Rule)
• Number of code words of length 3 263
  (Multiplication Rule)
• Total number of code words 26 262 263
  (Addition Rule)

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1. Basic Rules Addition/Difference Rule

• Example 2 There are 15 different computer
  science books, 12 different math books, and 10
  different chemistry books on the shelf. How many
  ways can we select 2 books, each from a different
  subject?

Answer

• By addition rule Set of selections of 2
  books Set of selection of 1 book from CS
  and 1 book from Math Set of selection of 1
  book from CS and 1 book from Chem Set of
  selection of 1 book from Math and 1 book from
  Chem

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1. Basic Rules Addition/Difference Rule

• Example 2 There are 15 different computer
  science boks, 12 different math books, and 10
  different chemistry books on the shelf. How many
  ways can we select 2 books, each from a different
  subject?

Answer
Set of selections of 2 books
1 book from CS and 1 book Math
1 book from CS and 1 book from Chemistry
1 book from Math and 1 book from Chemistry
15 12 (M.R.)
15 10 (M.R.)
12 10 (M.R.)
Addition rule is reasoning-by-cases applied to
counting!
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1. Basic Rules Addition/Difference Rule

• Example 3 A group of eight people are attending
  the movies together. If two of the eight people
  are enemies and do not want to sit next to each
  other, how many ways can this group sit in a row,
  such that the two enemies are separated?

Answer
Different arrangements of 8 people in a row
8! (by multiplicaton rule) How? Step 1 Sit
the 1st person 8 ways Step 2 Sit the 2nd
person 7 ways regardless of outcome of step
1. Step 3 Sit the 3rd person 6 ways regardless
of outcome of step 1-2.
Different arrangements of 8 people in a row where
the 2 enemies sit next to each other
Different arrangements of 8 people in a row where
the 2 enemies sit apart
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1. Basic Rules Addition/Difference Rule

• Example 3 A group of eight people are attending
  the movies together. If two of the eight people
  are enemies and do not want to sit next to each
  other, how many ways can this group sit in a row,
  such that the two enemies are separated?

Answer
Different arrangements of 8 people in a row
8!
27! (by multiplicaton rule) How? Step 1
Sit the 2 enemies together 27 ways Step 2-7
Sit the remaining 6 people 654321
Different arrangements of 8 people in a row where
the 2 enemies sit next to each other
Different arrangements of 8 people in a row where
the 2 enemies sit apart
OR Another way of looking at it
Step 1 Combine the 2 enemies as 1 person and
take it as an arrangement of 7 people to 7
chairs, (7!). Step 2 Different ways of arranging
the 2 enemies to sit side-by-side 2 ways.
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1. Basic Rules Addition/Difference Rule

• Example 3 A group of eight people are attending
  the movies together. If two of the eight people
  are enemies and do not want to sit next to each
  other, how many ways can this group sit in a row,
  such that the two enemies are separated?

Answer
Different arrangements of 8 people in a row
8!
27! (by multiplicaton rule)
Different arrangements of 8 people in a row where
the 2 enemies sit next to each other
Different arrangements of 8 people in a row where
the 2 enemies sit apart
Answer 8! 27! (Difference Rule)
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1. Basic Rules Addition/Difference Rule

• Example 4 How many integers are there in 1000 to
  9999 that contain at least a digit 5.

Answer Number of integers in 1000 to 9999
Number of integers in 1000 to 9999 that do not
contain a digit 5. (Difference Rule) (9999
1000 1) 8.93 (Linear Series Rule)
(Multiplication Rule) 2439
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1. Basic Rules Addition/Difference Rule

• Example 5 How many 3 digit numbers have at least
  one digit repeated?

Answer Number of 3 digit numbers Number
of 3 digit numbers which have NO digit repeated
(Difference Rule)
(9 10 10) Multiplication Rule Step
1 Choose hundreths digit (must exclude leading
0, therefore only 9 ways) Step 2 Choose
tenths digit Step 3 Choose units digit
(9 9 8) Multiplication Rule Step
1 Choose hundreths digit (must exlude leading
0) Step 2 Choose tenths digit (10 ways,
excluding the digit in step 1. Therefore 9
ways) Step 3 Choose units digit
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1. Basic Rules Inclusion-Exclusion Rule

• 1.4 Inclusion-Exclusion Rule
• (For 2 sets) Given any sets A and B, A È B
  A B A Ç B
• (For 3 sets) Given any sets A, B and C, A È B
  È C A B C A Ç B A Ç C B
  Ç C A Ç B Ç C

• (For n sets) Given any sets A1 An, A1 È
  È An Ai
• Ai Ç Aj
• Ai Ç Aj Ç Ak

i Î 1..n
i,j Î 1..n i,j distinct
i,j,k Î 1..n i,j,k distinct
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1. Basic Rules Inclusion-Exclusion Rule

• Moral of the story behind the inclusion-exclusion
  rule
• If you over-count, then subtract away those you
  counted more than once.
• If you under-count, then add back those you have
  missed out.
• This is applicable to any counting scenario.
• Inclusion-Exclusion Rule is the generalized
  version of the addition rule. If the sets are
  disjoint, then the In-Ex rule reduces to the
  addition rule.

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1. Basic Rules Inclusion-Exclusion Rule

• Example 1 How many integers from 1 through 1000
  are multiples of 3 or 5? How many are neither
  multiples of 3 nor 5?

Answer Integers of multiples of 3 or
5 Integers of multiples of 3
Integers of multiples of 5 Integers
of multiples of 3 and 5 (Inclusion-Exclusio
n Rule)
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1. Basic Rules Inclusion-Exclusion Rule

• Example 1 How many integers from 1 through 1000
  are multiples of 3 or 5? How many are neither
  multiples of 3 nor 5?

Answer Integers of multiples of 3 or
5 Integers of multiples of 3
Integers of multiples of 5 Integers
of multiples of 3 and 5 (Inclusion-Exclusio
n Rule)
Integers of multiples of 3 3k where k
1,2,333 333 Integers of multiples of 5
5k where k 1,2,200 200 Integers of
multiples of 3 and 5 15k where k
1,2,66 66 Ans 333 200 66 467
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1. Basic Rules Inclusion-Exclusion Rule

• Example 1 How many integers from 1 through 1000
  are multiples of 3 or 5? How many are neither
  multiples of 3 nor 5?

Question How many are neither multiples of 3 nor
5? Answer Int from 1..1000 with are multiples
of 3 or 5 Í Int from 1..1000
Ans 1000-467 (By difference rule) 533
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1. Basic Rules Inclusion-Exclusion Rule

• Example 2 In a class of 50 students, 30 know
  Pascal, 18 know Fortran, 26 know COBOL, 9 know
  both Pascal and Fortran, 16 know Pascal and
  COBOL, 8 know both Fortran and COBOL, 47 knows at
  least one of the three languages.
• (a) how many students know none of the three
  languages?
• (b) how many students know all three languages?
• (c) how many students know Pascal and Fortran
  but not COBOL?
• (d) how many students know Pascal, but neither
  Fortran nor COBOL?

P set of students who know Pascal C set of
students who know COBOL F set of students who
know Fortran U All 50 students.
(a) U (P È C ÈF) U (P È C
ÈF) (By difference rule, since (P È C ÈF)
Í U) U 50 (Given) (P È C ÈF) 47
(Given) Therefore answer 50-47 3.
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1. Basic Rules Inclusion-Exclusion Rule

• Example 2 In a class of 50 students, 30 know
  Pascal, 18 know Fortran, 26 know COBOL, 9 know
  both Pascal and Fortran, 16 know Pascal and
  COBOL, 8 know both Fortran and COBOL, 47 knows at
  least one of the three languages.
• (a) how many students know none of the three
  languages?
• (b) how many students know all three languages?
• (c) how many students know Pascal and Fortran
  but not COBOL?
• (d) how many students know Pascal, but neither
  Fortran nor COBOL?

P set of students who know Pascal C set of
students who know COBOL F set of students who
know Fortran U All 50 students.
(b) (P È C ÈF) P C F PÇC
PÇF CÇF PÇCÇF (Inclusion-Exclusio
n Rule for 3 sets) 47 30 26 18 9 16
8 PÇCÇF PÇCÇF 6
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1. Basic Rules Inclusion-Exclusion Rule

• Example 2 In a class of 50 students, 30 know
  Pascal, 18 know Fortran, 26 know COBOL, 9 know
  both Pascal and Fortran, 16 know Pascal and
  COBOL, 8 know both Fortran and COBOL, 47 knows at
  least one of the three languages.
• (a) how many students know none of the three
  languages?
• (b) how many students know all three languages?
• (c) how many students know Pascal and Fortran
  but not COBOL?
• (d) how many students know Pascal, but neither
  Fortran nor COBOL?

P set of students who know Pascal C set of
students who know COBOL F set of students who
know Fortran U All 50 students.
(c) ((PÇF) (PÇFÇC)) PÇF
PÇFÇC (Difference Rule) 9 6 3
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1. Basic Rules Inclusion-Exclusion Rule

• Example 2 In a class of 50 students, 30 know
  Pascal, 18 know Fortran, 26 know COBOL, 9 know
  both Pascal and Fortran, 16 know Pascal and
  COBOL, 8 know both Fortran and COBOL, 47 knows at
  least one of the three languages.
• (a) how many students know none of the three
  languages?
• (b) how many students know all three languages?
• (c) how many students know Pascal and Fortran
  but not COBOL?
• (d) how many students know Pascal, but neither
  Fortran nor COBOL?

P set of students who know Pascal C set of
students who know COBOL F set of students who
know Fortran U All 50 students.
(d) Similar to (c), those who know Pascal and
COBOL, but not Fortran 16 6 10.
P
C
10
?
Since P 30 By difference rule, those who
know Pascal, but neither fortran nor COBOL 30
19 11
6
3
F
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Summary so far Basic Counting Rules

• Four basic Rules
• Linear Series Rule
• Multiplication Rule
• Addition/Difference Rule
• Inclusion-Exclusion Rule
• All counting can be broken down to these four
  rules they may be likened to the atoms of
  counting.
• IMPT Just like the logical inference rules, you
  use and combine them together to create a proof,
  also here, you have to combine the use of the
  counting rules when necessary, to tackle the
  overall counting problem
• We now use our atoms to build common
  procedural calls to tackle common counting
  scenarios
• Permutations
• Combinations

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• End of Lecture