Lecture 4 part 2

1
Lecture 4 (part 2)

• Combinatorics
• Reading Epp Chp 6

2
Outline

• Basic Rules
• Linear Series Rule
• Multiplication Rule
• Addition and Difference Rule
• Inclusion-Exclusion Rule
• Common Counting Scenarios
• Permutations (Ordered Selections)
• Combinations (Unordered Selections)
• Permutations with repetitions
• Combinations with repetitions
• Algebra of Combinations
• Binomial Theorem

3
2.1 Permutations

• Definition
• An r-permutation of a set of n elements is an
  ordered selection of r elements taken from the
  set of n elements (1 r n).
• Notation
• The number of r-permutations of a set of n
  elements is denoted as P(n,r).
• Note permutations are ordered selections
  selecting the object in the first try/position,
  is considered different from selecting that
  object in the second or subsequent
  tries/positions.

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2.1 Permutations

• Theorem
• If n and r are integers such that 1 r n, then
• P(n,r) n(n-1)(n-2)(n-r1)
• or equivalently
• P(n,r) n! / (n-r)!
• Proof
• Uses the multiplication rule.
• Formal proof is by induction. Informal proof is
  as follows

1 r n
Number of choices in step i is n-(i-1),
regardless of the choices made in previous steps
gt Can use Multiplication Rule
n(n-1)(n-2)(n-r1)
5
2.1 Permutations

• Theorem
• If n and r are integers such that 1 r n, then
• P(n,r) n(n-1)(n-2)(n-r1)
• or equivalently
• P(n,r) n! / (n-r)!
• Corollary P(n,n) n! / (n-n)! n!

• Theorem CIRCULAR PERMUTATIONS
• The number of ways of permuting r objects from a
  set of n objects in a circle (in which two
  arrangements are the same when one is a rotation
  of the other) is P(n,r) / r
• When n r, we have P(n,n) / n (n-1)!

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2.1 Permutations

• Example 1 (a) How many different ways can three
  of the letters of the word BYTES be chosen and
  written in a row? (b) How many different ways
  can this be done if the first letter must be B

• Answer
• (a) The problem is a 3-permutation from a set of
  5 objects.
• P(5,3) 5!/2! 60
• (b) 2 step process (Multiplication Rule)
• Step 1 Assign B to the first position 1 way
• Step 2 Assign 4 letters to the remaining 2
  positions P(4,2) ways, regardless of step 1.
• Answer 1 x P(4,2) 12.

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2.1 Permutations

• Example 2
• (a) In how many ways can the letters in the word
  COMPUTER be arranged in a row?
• (b) In how many ways can the letters in the word
  COMPUTER be arranged in CO must appear together
  in order?

Answer (a) P(8,8) 8! (permutate all eight
objects) (b) P(7,7) 7! (Treat CO as a unit)
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2.1 Permutations

• Example 3
• (a) How many ways can a group of 6 people be
  seated around a table?
• (b) How many ways can a group of 6 people be
  seated around a table when two of them cannot sit
  next to each other?

Answer (a) P(6,6)/6 5! (circular
permutation) (b) By difference rule Number of
ways a group of 6 can sit around a table
Number of ways in which the 2 enemies sit next
to each other 5! (2xP(5,5)/5) (treat
the 2 enemies as 1 person) 5! (2x4!)
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2.2 Combinations

• Definition
• An r-combination of a set of n elements is a
  subset of r elements taken from the set of n
  elements (r n).
• Notation
• The number of r-permutations of a set of n
  elements is denoted as C(n,r), also as

• Note A combination is an unordered selection
  you are selecting a set, and ordering is not
  important in sets.

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2.2 Combinations

• Theorem
• If n and r are integers such that r n, then
• C(n,r) P(n,r) / r!
• or equivalently
• C(n,r) n! / ( r!(n-r)! )
• Informal Proof

Step 1 Ordered selection P(n,r)
Step 2 Get rid of duplicates since ordering is
not important P(n,r) / r!
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2.2 Combinations

• Example 1
• You are to select five members from a group of
  twelve to form a team.
• (a) How many distinct five-person teams can be
  selected?
• (b) If two of them insist on working together as
  a pair, such that any team must either contain
  both of neither. How many five person teams can
  be formed?

(b) Use Addition Rule
Answer (a) C(12,5)
All 5-person teams satisfying the 2-friends
constraint.
Those teams which involve the 2 friends
Those teams which do not involve the 2 friends
Step 1 select the two friends 1 way

C(10,5)
Step 2 select the remaining 3 people C(10,3)
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2.2 Combinations

• Example 2
• You are to select five members from a group of
  twelve to form a team.
• If two of them insist on working apart,how many
  five person teams can be formed?

Answer 1st version, using difference rule.
Those teams with the two enemies together
Those teams with the two enemies apart
Step 1 select the two enemies 1 way
Ans C(12,5) C(10,3)
Step 2 select the remaining 3 people C(10,3)
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2.2 Combinations

• Example 2
• You are to select five members from a group of
  twelve to form a team.
• If two of them insist on working apart,how many
  five person teams can be formed?

Answer 2nd version, using addition rule.
Those teams which contain A but not B
Those teams which contain B but not A
Those teams which do not contain A nor B
Step 1 select the A 1 way


Step 2 select the remaining people except B
C(10,4)
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2.2 Combinations

• Example 3
• A group of twelve consists of five men and seven
  women.
• (a) How many five-person teams can be chosen
  that consists of three men and two women?
• (b) How many five-person teams contain at least
  one man?
• (c) How many five-person teams contain at most
  one man?

(a) Answer Step 1 choose the men C(5,3)
ways Step 2 choose the women C(7,2) ways
(regardless of the choices made in step
1) Multiplication rule C(5,3) x C(7,2)
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2.2 Combinations

• Example 3
• A group of twelve consists of five men and seven
  women.
• (a) How many five-person teams can be chosen
  that consists of three men and two women?
• (b) How many five-person teams contain at least
  one man?
• (c) How many five-person teams contain at most
  one man?

(b) Answer Number of 5-person teams that
contain at least one man Number of 5-person
teams Number of 5-person teams that do not
contain any men (all women). (DIFFERENCE
RULE) C(12,5) C(7,5)
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2.2 Combinations

• Example 3
• A group of twelve consists of five men and seven
  women.
• (a) How many five-person teams can be chosen
  that consists of three men and two women?
• (b) How many five-person teams contain at least
  one man?
• (c) How many five-person teams contain at most
  one man?

(b) Answer Number of 5-person teams that
contain at most one man Number of 5-person
teams that contain no men Number of 5-person
teams that contain 1 man (BY ADDITION RULE)
(C(5,0) x C(7,5)) Step 1 Choose 0 men from 5
men Step 2 Choose 5 women from 7 women
(C(5,1) x C(7,4)) Step 1 Choose 1 man from 5
men Step 2 choose 4 women from 7 women
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2.2 Combinations

• Example 4
• 10 people are to sit around two tables in a
  hawker (dealer) centre. The first table has 6
  chairs, the second table has 4 chairs.
• (a) How many ways can this be done?
• (b) How many ways can this be done if two of
  them needs to sit together?

(a) Answer (version1) Step 1 Choose 6 people
to sit on the first table C(10,6) Step 2
Arrange the 6 people on the first table 5!
(Circular Permutation) Step 3 Arrange the
remaining 4 on the second table. 3! (Circular
Permutation) (BY MULTIPLICATION RULE)
C(10,6) x 5! x 3! 10!/24
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2.2 Combinations

• Example 4
• 10 people are to sit around two tables in a
  hawker centre. The first table has 6 chairs, the
  second table has 4 chairs.
• (a) How many ways can this be done?
• (b) How many ways can this be done if two of
  them needs to sit together?

(a) Answer (version2) Step 1 Permute 6 people
from 10 circularly around the first
table P(10,6)/6 Step 2 Arrange the remaining
4 on the second table. 3! (Circular
Permutation) (BY MULTIPLICATION RULE)
P(10,6)/6 x 3! 10!/24
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2.2 Combinations

• Example 4
• 10 people are to sit around two tables in a
  hawker centre. The first table has 6 chairs, the
  second table has 4 chairs.
• (a) How many ways can this be done?
• (b) How many ways can this be done if two of
  them needs to sit together?

(b) Answer (version1) BY ADDITION RULE Number
of sitting arrangements where the two sit on the
1st table Number of sitting arrangements
where the two sit on the 2nd table
BY MULTIPLICATION RULE Step 1 Put the two on
the first table 1 way Step 2 Select 4 more to
join them C(8,4) Step 3 Permute them around
the table 5! Step 4 Permute the remaining 4
circularly on 2nd table 3!
BY MULTIPLICATION RULE Step 1 Put the two on
the second table 1 way Step 2 Select 2 more
to join them C(8,2) Step 3 Permute them
around the table 3! Step 4 Permute remaining
6 circularly on 1st table 5!
(C(8,4)x5!x3!) (C(8,2)x3!x 5!)
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2.2 Combinations

• Example 4
• 10 people are to sit around two tables in a
  hawker centre. The first table has 6 chairs, the
  second table has 4 chairs.
• (a) How many ways can this be done?
• (b) How many ways can this be done if two of
  them needs to sit together?

(b) Answer (version2) BY ADDITION RULE Number
of sitting arrangements where the two sit on the
1st table Number of sitting arrangements
where the two sit on the 2nd table
BY MULTIPLICATION RULE Step 1 Put the two on
the first table 1 way Step 2 Permute 4 from 8
circularly on the 2nd table P(8,4)/4 Step 3
Permute the 6 circularly on 1st table 5!
BY MULTIPLICATION RULE Step 1 Put the two on
the second table 1 way Step 2 Permute 6 from
8 circularly on the 1st table P(8,6)/6 Step 3
Permute remaining 4 circularly on 2nd table 3!
(P(8,4)/4 x 5!) (P(8,6)/6 x 3!)
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2.3 Permutations with repetitions

• Theorem (Permutations from a multi-set)
• Given a collection of n objects of which
• n1 are of type 1 and are indistinguishable from
  each other
• n2 are of type 2 and are indistinguishable from
  each other
• nk are of type k and are indistinguishable from
  each other
• and that n1 n2 nk n, then the number
  of distinct permutations of the n objects is
• C(n, n1) C(n n1 , n2) C(n n1 n2 , n3)
  C(n n1 n2 nk-1 , nk)
• or equivalently
• n! / (n1! n2! nk! )

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2.3 Permutations with repetitions

• Example 1
• How many 8-bit strings have exactly three 1s?

• Answer (version1 details)
• Example
• 10001101 is an 8-bit string, but has four 1s
• 10010001 is an 8-bit string and has exactly three
  1s.
• TECHNIQUE Sometimes, you have to know (a) the
  set you are choosing from and (b) the
  destination type that you are choosing to.
• Point of view1 Choose from the set 0,1 and
  place into the bit positions 1,2,3,4,5,6,7,8.
• Point of view2 Choose from the bit positions
  1,2,3,4,5,6,7,8, and place them into the set
  0,1 (i.e. assign them to either a bit 0 or 1)
• Adopting different points of view of the same
  problem may help.
• Use point of view2
• Step 1 Choose 3 bit positions from 8 to be
  assigned a bit 1 C(8,3)
• Step 2 Choose the remaining 5 bit positions for
  bit 0 C(5,5) 1
• Answer C(8,3) (By multiplication rule)

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2.3 Permutations with repetitions

• Example 1
• How many 8-bit strings have exactly three 1s?

• Answer (version2 direct application of
  formula)
• What are the objects that you are permuting
• Point of view1 1s and 0s with duplicates?
• Point of view2 the bit positions
• Again, it is still using point of view2.
• Permute 8 objects (the bit positions)
• 3 of which are indistinguishable from each other
• 5 of which are indistinguishable from each other
• Answer 8!/(3! x 5!)

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2.3 Permutations with repetitions

• Example 2
• How many ways are there of ordering the letters
  of the word MISSISSIPPI, which are
  distinguishable from each other?

• Answer
• What are the objects that you are permuting
• Point of view1 The Letters M,I,S,P? (with
  duplicates?)
• Point of view2 The 11 positions where the
  letters can be placed
• Again, point of view2.
• Permute 11 objects (the letter positions)
• 4 of which are indistinguishable from each other
  (for the letter S)
• 4 of which (a different type) are
  indistinguishable from each other (for the letter
  I)
• 2 of which (a different type) are
  indistinguishable from each other (for the letter
  P)
• 1 position for letter M
• Answer (use formula) 11!/(4! x 4! x 2! x 1!)
• Answer (from scratch) C(11,4) x C(7,4) x C(3,2)
  x C(1,1)

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2.4 Combinations with repetitions

• Definition (r-combinations from a multi-set)
• Given a set X of n objects, an r-combination
  with repetition allowed (or r-combination with a
  multi-set of size r) is an unordered selection of
  elements taken from X with repetition allowed.
• Theorem The number of r-combination with
  repetition allowed drawn from a set of n elements
  is C(rn1 , r).

• Example 3-combinations from a set a,b,c,d
• a,a,a a,a,b a,a,c a,a,d
• a,b,b a,b,c a,b,d
• a,c,c a,c,d a,d,d
• b,b,b b,b,c b,b,d
• b,c,c b,c,d b,d,d
• c,c,c c,c,d c,d,d
• d,d,d
• Total Number 20 C(34-1 , 3)

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2.4 Combinations with repetitions

• Definition (r-combinations from a multi-set)
• Given a set X of n objects, an r-combination
  with repetition allowed (or r-combination with a
  multi-set of size r) is an unordered selection of
  elements taken from X with repetition allowed.
• Theorem The number of r-combination with
  repetition allowed drawn from a set of n elements
  is C(rn1 , r).

• Example 3-combinations from a set a,b,c,d

a,a,a
X X X
a,a,b
a,b,d
a,c,d
b,c,c
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2.4 Combinations with repetitions

• Definition (r-combinations from a multi-set)
• Given a set X of n objects, an r-combination
  with repetition allowed (or r-combination with a
  multi-set of size r) is an unordered selection of
  elements taken from X with repetition allowed.
• Theorem The number of r-combination with
  repetition allowed drawn from a set of n elements
  is C(rn1 , r).

• Example 3-combinations from a set a,b,c,d

a,a,a
X X X
a,a,b
X X
X
a,b,d
X
X
X
a,c,d
X
X
X
b,c,c
X
X X
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2.4 Combinations with repetitions

• Definition (r-combinations from a multi-set)
• Given a set X of n objects, an r-combination
  with repetition allowed (or r-combination with a
  multi-set of size r) is an unordered selection of
  elements taken from X with repetition allowed.
• Theorem The number of r-combination with
  repetition allowed drawn from a set of n elements
  is C(rn1 , r).

• Example 3-combinations from a set a,b,c,d

• Problem generalized
• r-combination putting r crosses
• From a set of n elements putting n-1 s in
  between crosses.
• Reduces to the same problem of assigning 2-bits
  (X and ) to rn-1 positions. (Permuting rn-1
  positions from a multi-set of X,

Selection
Represented by
a,a,a
a,a,b
a,b,d
a,c,d
b,c,c
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2.4 Combinations with repetitions

• Example 1 A person giving a party wants to set
  out 15 assorted cans of soft drinks for his
  guests. He shops at a store that sells 5
  different types of soft drinks
• (a) How many different selections of cans of 15
  soft drinks can he make?
• (b) If root beer is one of the types of soft
  drinks, how many different selections include at
  least 6 cans of root beer?

• Answer
• (a) 15-combination (15 drinks) from a multi-set
  of 5 categories
• r 15 drinks 15 crosses
• n 5 categories need 4 to separate the
  crosses
• C(155-1 , 15) C(19, 15) 3876
• (b) Step 1 take out 6 cans of root beer 1 way
• Step 2 select the remaining 9 cans
  9-combination (9 remaining drinks) from a
  multi-set of 5 categories C(95-1 , 9) C(13,9)
  715
• Final Answer 1 x 715 715 (Multiplication
  Rule)

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2.4 Combinations with repetitions

• Example 2 How many solutions are there to the
  equation
• x1 x2 x3 x4 10
• (a) if x1, x2, x3, x4 are non-negative integers?
• (b) if x1, x2, x3, x4 are positive integers?

• Answer
• (a) 10-combination from a multi-set of 4
  categories
• r 10 units 10 crosses to be distributed
  into
• n 4 categories the four variables need 3
  to separate the crosses
• C(104-1 , 10) C(13, 10) 286
• (b) Step 1 Assign 1 unit to each of the four
  variables 1 way.
• Step 2 Assign the remaining 6 units to the four
  variables
• 6-combination from a multi-set of 4 categories
• C(64-1 , 6) C(9,6) 84
• Final Answer 1 x 84 84 (Multiplication Rule)

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2.4 Combinations with repetitions

• Example 3 How many triplets of the form (i,j,k)
  are there, when
• (a) each i , j , k Î 1,,n
• (b) each i , j , k Î 1,,n and 1 i j k n

• Answer
• (a) Permutation n3
• (b) 3-combinations from a multi-set of n
  categories
• (Observation skills needed to relate problem to
  known scenario)
• For example if 1 i j k 5, then we have
  3-combinations from a multi-set of 5 categories
  3 Xs and 4 s
• (1,1,2) being represented as XXX
• (1,2,4) being represented as XXX
• (2,3,5) being represented as XXX
• Therefore in general, answer is C(n3-1 , 3)
• (n2)!/ (3! x (n-1)!)
• (n2)(n1)n / 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)
• Permutations with repetitions
• Combinations with repetitions
• Algebra of Combinations
• Binomial Theorem

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3. Algebra of Combinations

• Common properties
• C(n,r) C(n, n-r)
• Pascals Formula C(n1, r) C(n, r-1) C(n,r)

• Proof of (1)
• Proven Algebraically
• C(n,r) n! / ( r! (n-r)! ) n! / ( (n-r)!
  (n-(n- r))! ) C(n, n-r)
• Proven using combinatorial reasoning (p331 of
  text)
• Let A be a set with n elements.
• Let the subsets of A of size r be B1, B2,,Bk.
• Each Bk can be paired up with a subset of A of
  size nr namely A Bk.
• All subsets of size n All subset of size nr
• B1 A B1
• B2 A B2
• Bk A Bk

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3. Algebra of Combinations

• Common properties
• C(n,r) C(n, n-r)
• Pascals Formula C(n1, r) C(n, r-1) C(n,r)

• Proof of (2) refer to p334 of recommended text.
• Implications Pascals Triangle

1
35
4. Binomial Theorem

• Binomial Theorem
• (ab)n an-kbk

• Algebraic proof is by induction on n. (p341 of
  text)
• Combinatorial Proof
• (ab)n (ab) (ab) (ab) (ab) (n copies)
• 1 2 3
  n
• b a a
  a
• a b a
  a
• What is the coefficient of an-kbk ?
• How many ways can we create the an-kbk term?
• Well, to create a term, we have to select either
  a or b from each (ab) group.
• Problem reduces to a n-permutation with
  repetitions (previous section 2.3), of 2 types of
  objects selecting n-k copies of a and k copies
  of b n!/(k!(n-k)!) C(n,k)

36
4. Binomial Theorem

• Binomial Theorem provides rapid expansion of
  polynomials.
• Example 1 Expand (ab)n.
• C(5,0)a5b0 C(5,1)a4b1 C(5,2)a3b2
  C(5,3)a2b3 C(5,4)a1b4 C(5,5)a0b5
• a5 5a4b 10a3b2 10a2b3 5ab4 b5
• It also allows you to access a particular term
  without the need to expand the entire polynomial
• Example 2 What is the coefficient of a9b6 in
  the expansion of (ab)15?
• Ans C(15,6)

37
4. Binomial Theorem

• Corollary of Binomial Theorem
• 2n (11)n

Q If S n, what is P(S)? A By addition
rule number of subsets of size 0 number
of subsets of size 1 number of subsets of
size 2 number of subsets of size n
38

• End of Lecture