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Chapter 6: Combinatorics
Discrete Math for CS
How to Count:

Combinatorics is about counting.

Pigeonhole was about counting too.

General Counting Problem:

considers the selection of items from a bigger set either with
or without repetition and with or without a specific order of
selection
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Example:

Problem 1: At the end of the day a small bakery has a number
of unsold cakes:


4 vanilla cakes,

2 chocolate cakes and

3 fruit cakes
If a customer comes in in a rush and orders 1 cake, how many
different choices does the customer have?
ANS: 4+2+3 = 9 choices
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Example:

Problem 2: A mixed-double team is to be chosen to represent a
local tennis club. There are:


6 male players and,

9 female players
How many different possible teams are there?
ANS: any one of 6 males and 9 females so 6x9 = 54 possible teams
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Addition Principle:

If A and B are disjoint events and n1 ways A can happen while
n2 ways B can happen then there are n1 + n2 ways either A or B
can happen.
Problem 1 is an example of the Addition Principle
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Multiplication Principle:

If there is a sequence of k events and n1 ways the first event
can happen, n2 ways the second event can happen,etc. and
finally nk ways the kth event can happen then the total number of
possible outcomes of all k events is are n1 x n2 x ... x nk.
Problem 2 is an example of the Mulitplication Principle
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Inclusion and Exclusion Principle:

From Chapter 3 we have the following formula
| A B | = | A | + | B | - | A B |

The Addition Principle is a special case of this formula where
the two sets are disjoint. Hence A B = 

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Multiplication Principle seen Mathematically:

Let Ai denote the ni outcomes of ith event, i = 1,...,k.

Then any sequence of k events is corresponds to an element of
the cartesian product
A1 x A2 x ... x Ak
so the number of different sequences is equal to the number of
elements in the cartesian product
|A1| x |A2| x ... x |Ak|
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Example:

Problem 3: How many 3-digit numbers begin with 3 or 4?:
ANS: Two disjoint cases: 3-digit numbers beginning with 3 and 3-digit numbers
beginning with 4. We use the Addition Principle to add the number of
outcomes of each case.
Case 1: there are 10 choices of the 2nd digit and 10 more choices of the 3rd so
100 choices of a 3-digit number beginning with 3. This is just the Multiplication
Principle.
Case 2: 100 choices of a 3-digit number beginning with 4.
A total of 100+100 = 200 different numbers.
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Example 1:

I take two pieces of fruit each day for lunch. I have 3 bananas, 4
apples and 2 pears at home. How many different ways can I
select two fruit for lunch?
ANS:
If I select a banana and apple I have 4 x 3 choices
If I choose a banana and pear I have 4 x 2 = 8 choices
If I choose an apple and a pear I have 3 x 2 = 6 choices.
I have 12 + 8 + 6 = 26 different ways to choose the fruit for lunch.
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Example 2:

How many different licence plates are there consisting of 6
characters, the first 3 characters are letters and the last 6 are
digits.
ANS:
I have 26 x 26 x 26 = 17,576 ways to choose the letters.
I have 10 x 10 x 10 = 1,000 ways to choose the digits.
I have 17,576 x 1,000 = 17,576,000 different possible licence plate
combinations.
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Counting Formulae:

A child has a bag of three kinds of candies Christmas Stripe(A),
butterscotch(B) and Hershey's(C) - . In how many ways can the
child select two candies.

Hard to answer this question because we don't know if it is
permitted to choose two of the same kind or if choosing a
butterscotch candy and then a Hershey's is the same as
choosing a Hershey's (BC) and then a butterscotch (CB)
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4 Cases

Repeats allowed and Order matters:
{AA, AB, AC, BA, BB, BC, CA, CB, CC} – total of 9 ways

Repeats allowed but Order does not matter:
{AA, AB, AC, BB, BC, CC} – total of 6 ways

Repeats not allowed but Order matters:
{ AB, AC, BC, BA, CA, BC} – total of 6 ways

Repeats not allowed and Order does not matter:
{ AB, AC, BC} – total of 3 ways
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How to Decide Which Case Applies?

Perhaps we are told that the child blindly puts her hand into a
bag containing the candy. This implies repeats are allowed but
really doesn't tell us if the girl cares about the order.

If we are told that she picks the candies out and gives the first
chosen candy to friend X and the second chosen candy to
friend Y then order clearly matters, at least to X and Y.
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Order Matters and Repetition Allowed:

We choose k items from a set of n items.



We call such a selection a k-sample.
Since repetition is allowed there are n ways to choose an item
each time a choice is made.
The Multiplication Principle tells us there are
n x n x ... x n = nk
possible k-samples.
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Example:

A computer represents integers using N binary digits.



The first digit indicates sign (+ or -).
The remaining N-1 digits represent the magnitude of the
integer.
How many distinct integers can be represented?
ANS: There are 2N different binary strings. Each one is a separate
integer except for 00...0 and 10...0. Both these have magnitude 0
and opposite signs so represnt the same integer; namely 0.
So there are 2N-1 different integers represented this way.
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k-permutations:

Select k objects from a set of n objects and repetitions are not
allowed and order matters.

Such a selection is called a k-permutation.
Example: You have n distinct things and each time you pick one it
is no longer available to be chosen again.
Pick 5 students from a class of 20 to become the class leaders.
How many different executive teams of students can you form?
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k-permutations:

P(n,k) = the number of ways of choosing k objects from a set of
n objects without repetition and order matters
There are n ways to chose the first item
For the second choice, the first-chosen item is missing so n-1
ways of choosing the second item.
For the third choice, the first two-chosen items are missing so
n-2 ways of choosing the third item.
P(n,k) = n(n-1)(n-2)...((n-k+1) = n!/(n-1)!
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Example:

How many four-letter words can you make from distinct letters
in the list a, g, m, o, p and r?
NOTE: Saying “distinct” means no repetitions. Order matters since a
different letter order gives a different word.
P(6,4) = 6!/(6-4)! = 6!/2! = 6 
5
4
Please ignore the text on this example.
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3 = 360
k-combination:

Select k objects from a set of n objects where order does not
matter and repetitions are not allowed.

Such a selection is called a k-combination.
Example: You have n things whose differences are not important to you
and each time you pick one it is no longer available to be chosen again.
Pick 5 students from a class of 20 to set up chairs for a school assembly.
How many different teams of students can you form?
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k-combinations:

C(n,k) = the number of ways of choosing k objects from a set of
n objects without repetition and order does not matter
If order matters we get the number P(n,k).
By the multiplication principle there are k! orderings of
each distinct choice of k elements. So the number of
k-permutations of k distinct objects selected from n objects
is the same as the number of unordered ways to select the
same objects times the number of ways to order the objects
chosen.
P(n,k) = C(n,k) 
k!
So C(n,k) = P(n,k) k!
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Example:

In a Chinese Restaurant you can order any three dishes for
what they call the 3-combination meal. How many different 3combination meals are there if there are 7 main dishes on teh
menu.
It doesn't make any difference what order you receive the dishes
in and we assume you must order three different dishes.
So order doesn't matter and there is selection without replacement.
C(7,3) = 7!/4!3! = 35
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k-selection:

We finally consider an unordered selection of k objects from a
collection of n objects with repetitions allowed.

This is called k-selection
Example: Since order is unimportant and repetitions are allowed we can
group all like-objects together.
Suppose we want to make a k-selection of 5 objects from the letters
a, b and c.
<group of as>|<group of bs>|<group of cs>
For example aa|b|cc or aaa||cc.
This reduces to finding out how many ways we can insert the two
markers, | and |, in a sequence of 7 slots. The other 5 slots are letters.
This number is C(7,2) = 21.
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Example:

Five dice are thrown. How many different outcomes are
possible if an outcome is a list of the upper faces in ascending
order.
Each dice has 6 outcomes. If 5 dice are thrown we have the result
<sequence of 1s>|<sequence of 2s>| ...|<sequence of 6s>
and we are really asking how many ways we can insert 5 markers
in 10 slots. For example 111||3||5|. This is C(10,5) = 252.
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k-selection formula:

The number of k-selections from n items is:
C(n+k-1,n-1) = (n+k-1)!/k!(n-1)!
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Summary:
Order Matters
k-sample: nk
Order Doesn't Matter
k-selection: (n+k-1)!/k!(n-k)!
k-permutation: n!/(n-k)!
k-combination: n!/k!(n-k)!
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Further Examples:

The key is to know what formula to use.

In a National Lottery a twice-weekly drawing takes place in
which 6 numbers are randomly drawn from the numbers 1..49.

Problem: Find the probability of winning.

Analysis: Without replacement and order does not matter.

Solution:
k-combination: C(49,6) ways to choose the numbers
Prob(winning) = 1/C(49,6) = 1/13,983,816
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Further Analysis:



Should I buy a ticket?
Statistically this equates to calculating the statistical outcome of
buying a ticket.
Simplify the situation and assume there is only one winner. The
initial lottery pot in NYS is about $3M.
Outcome = amount won * prob of winning + amount lost * prob of losing
= 3,000,000 * 1/13,983,816 + -1 * 13,983,815/13,983,816
= -0.79
So the outcome is negative and this is not a good bet.
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Further Analysis:

What if you win $10 if you get three numbers out of the 6? What
is the probability of this happening?
Prob(guessing 3 numbers correctly) =
number of ways of guessing three correct numbers/
number of ways of guessing.
Guessing 3 correct numbers from the 6 correct numbers can
be done C(6,3) ways – no repeats, order unimportant
Guessing three incorrect numbers can be done C(43,3) ways.
Total number of winning combinations is:
C(6,3) x C(43,3) = 246,820
Prob(winning something) = 246,820/13,983,816 = 0.017
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Example:

Twelve people, including Peter and Mary are candidates for a
committee of 5. How many different committees are possible?

including Peter and Mary,

excluding Peter and Mary,

containing either Peter and Mary but not both?
Answers:
How many committees? No replacement, order unimportant – C(12,5) = 792.
Including Peter and Mary? Peter and Mary are selected so we need to choose
three more from 10 people – C(10,3) = 120
excluding Mary and Peter? Select 5 from 10 – C(10,5) = 252
either Peter or Mary? Mary only – C(10,4). Peter only – C(10,4)
either Mary or Peter = 2 x C(10,4) = 420
Alternatively, the 792 possible committees fall into one
of the above categories. So the third category has
792 – 120 – 252 = 420 members.
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Binomial Expansion:

The numbers C(n,k) are the coefficients of the expansion of
(a+b)n.
Example: (a+b)3 = (a+b)(a+b)(a+b) = a3 + 3a2b + 3ab2 + b3
and C(3,0) = 1, C(3,1) = 3, C(3,2) = 3 and C(3,3) = 1


This turns out to be true because when we multiply out
(a+b)(a+b)(a+b) we are essentially selecting either an a or a b
from each term (a+b) and the coefficient of akbn-k is the number
of ways of selecting k a's from n without repetition and ignoring
order.
Of course, 0! = 1 by definition.
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(a+b)n:
(a+b)n = C(n,0)an + C(n,1)an-1b + C(n,2)an-2b2 + ... + C(n,n-1)abn-1 + C(n,n)bn


The above is called the binomial expansion of (a+b)n.
Each C(n,k) is called a binomial coefficient.
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Pascal's Triangle:
1
1
1
1
1
1
2
3
4
1
3
6
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1
4
1
Pascal's Triangle:
1
1
1
1
1
1
2
3
4
1
3
6
C(n,k) = C(n-1,k-1) + C(n-1,k)
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1
4
1
Theorem:
C(n,k) = C(n-1,k-1) + C(n-1,k).
Proof: C(n-1,k-1) + C(n-1,k) = (n-1)!/(k-1)!(n-k)! + (n-1)!/k!(n-k-1)!
= (n-1)!/(k-1)!(n-k-1)! x (1/(n-k) + 1/k)
= (n-1)!/(k-1)!(n-k-1)! x n/((n-k)k)
= n!/(n-k)!k! = C(n,k)
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Final Topic:

How many arrangements can be made of objects (possibly
repeated) from a given set?
Example: Consider the word “defender”. This contains 4 distinct letters,
some repeated more than once.
We can arrange these letters 8! ways, but then we will see some are
repetitions since we can't tell one from another., for example.
Since there are 2 d's there are 2! ways of arranging these.
Since there are 3 e's there are 3! ways of arranging these.
There are a total of 8!/2!3! distinct orderings of the letters of “defender”.
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Rearrangement Theorem:
There are n!/(n1!n2!n3!...nk!) ways of rearranging a collection of n
objects where there are n1 identical objects of one type, n2
identical objects of another type, ..., and nk identical objects of a
final type. Clearly n = n1 + n2 + ... +nk.
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Example:

How many ways can we divide 15 students into 3 groups of 5
so that each group studies a different topic?

We have 15 objects divided into 3 identical groups of 5 each.

This can be done 15!/(5!5!5! ) = 68,796 ways.



Alternatively, imagine lining the 15 people up so that the first 5
go into group 1, the next 5 into group 2 and the last 5 into group
3.
For each distinct set of the first 5 students there are 5! different
orders we can ignore; likewise for the other two groups.
So there are 15!/(5!5!5!) different group assignment
possibilities.
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Multinomial:

The term n!/(n1!n2!n3!...nk!) is called a multinomial.

It is the form of the coefficients of the expansion of
(x1 + x2 + ... + xk)n


This follows since the term x1n1x2n2...xknk gets its coefficient by
selecting n1 instances of x1, n2 instances of x2, etc from the
bracketed terms (x1 + x2 + ... + xk).
This is the same as arranging n objects, n1 of type x1, n2 of type
x2, etc.
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Example:

Find the coefficient of x3y2z4 in the expansion of (x+y+z)9.
9!/(3!2!4!) = 1260

Find the coefficient of x3y2 in the expansion of (x+y+3)7.
If this were (x+y+z)7, then the coefficient of x3y2z2 would be
7!/(3!2!2!) = 210.
Since z = 3, the coefficient is 210x32. = 1890.
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Algorithm Efficiency:


Time and Space efficiency of a computer algorithm is a big
deal.
We often measure these aspects of an algorithm by counting
how much work is done or how much time is consumed as a
function of n, the size of the initial problem.
Example: Find if a particular word, X, is in a dictionary of size n.
Sequential Search:
This search compares X with the first word, the second word and so on
until either it is found or not found. The “worse case” requires n comparisons –
not there and X starts with a lot of zzz.
Binary Search:
Compare X to the middle word and use lexicographical ordering of words to
decide whether to continue the search in the first half or last half of the
dictionary. This method requires 1 + log2n comparisons.
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Example:

Suppose 5 algorithms – A, B, C, D and E – involve n, 3n2,
2n2+4n, n3 and 2n elementary operations. If each operation
takes 1 millisecond estimate the running times for n = 1, 10, 100
and 1000.
1
10
100
1000
A
n
1 ms
10 ms
100 ms
1000 ms
linear time
B
3n
3 ms
300 ms
30 sec
0.83 hr
2
C
D
2n2+4n
n3
6 ms
1 ms
240 ms
1 sec
20.4 sec 0.28 hr
0.56 hr 11.6 days
polynomial time
E
log2n
2 ms
1.024 sec
4x107 cent
101 7 6 cent
exponential time
Polynomial time algorithms with the same degree are
in the same ballpark.
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Time Complexity Functions:



Suppose f(n) and g(n) measure the time complexity of two
algorithms.
We say f(n) is of order at most g(n) (written O(g(n))) if there
exists a positive constant C such that |f(n)| <= C|g(n)| for all but
a finite number of distinct values of n.
Colloquially we say “f(n) is big-O of g(n)”.
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Example:

Show 2n2+4n is O(n2).
n <= n2 for n >= 1.
2n2+4n <= 2n2+4n2 = 6n2. for n >= 1
So letting C=6 in the above definition we can conclude
2n2+4n is O(n2)

Note: It is also the case that n2 <= 2n2+4n for n>=1 so we can
say n2 is O(2n2+4n) or that these two time complexity functions
have the same order of magnitude.
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Order of Magnitude Hierarchy


There is a hierarchy of functions, each of which is a greater
order of magnitude than its predecessors.
There are actually many such hierarchies but one is:
1
log n
n n2 n3 n4 ... nk ... 2n
Exercise: Where does nlogn (n(log n)) fit in?
As n increases the values of the above time complexity functions increase
more rapidly on the right than on the left.
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2048
1024
512
256
128
64
32
16
8
4
2
1
n
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Example:


A function like f(n) = 9n+3n6+7log n is assigned the time
complexity of O(n6) because 9n is O(n), 3n6 is O(n6) and 7log n
is O(log n). Since n and 7log n occur earlier in the hierarchy
than n6 we can say both n and 7log n are O(n6).
Since all three terms of the function f(n) are O(n6) the function
itself is O(n6).
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What to Count:

Consider the following pseudocode:
begin
for i:= 1 to 2n do
for j:= 1 to n do
for k := 1 to j do
x := x + 1
end
Determine the time complexity by counting the number of times
the assignment statement executes.
The outer loop executes 2n times.
The loop indexed by j executes n times for each i
For each value of j the statement x := x+1 executes j times
So for each value of i, the statement x := x+1 executes
1 + 2 + ... + n = 1/2(n(n+1)) times.
So the time complexity of the loops is
T(n) = 2n(1/2(n(n+1))) = n2(n+1)
So T(n) is O(n3).
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Pascal's Triangle:
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