Download Lecture 15 - Mathematics

Math236
Discrete Maths with Applications
P. Ittmann
UKZN, Pietermaritzburg
Semester 1, 2012
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The Multiplication Principle
Theorem
Let S be a set of k-tuples (s1 , s2 , . . . , sk ) of objects in which:
the rst object s1 comes from a set of size n1
for each choice of s1 there are n2 choices for object s2
for each choice of s2 there are n3 choices for object s3
for each choice of s3 there are n4 choices for object s4
and, in general, for each choice of si , 1 ≤ i ≤ k − 1, there are ni +1
choices for object si +1
Then the number of k-tuples in the set S is n1 n2 · · · nk
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The Multiplication Principle (cont.)
Example
How many 4-digit odd numbers are there?
We consider each number
abcd
as the 4-tuple
(a, b, c , d )
Such a number is odd if and only if the last digit,
d,
is in the set
{1, 3, 5, 7, 9}
There are no other restrictions on the digits
Thus:
n4 = 5
n1 = 9
(since the rst digit cannot be 0),
n2 = n3 = 10,
and
It follows that the number of 4-digit odd numbers is
9
· 10 · 10 · 5 = 4500
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Basic counting principles
Example
How many odd numbers less than 10,000 are there?
We use the Addition Principle together with the Multiplication
Principle
For
i ∈ {1, 2, 3, 4},
So,
N1 = {1, 3, 5, 7, 9}, N2 = {11, 13, 15, 17, 19, 21, . . . , 99}
Notice that
let
Ni
be the set of all
{N1 , N2 , N3 , N4 }
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i -digit
odd numbers
and so on
is a pairwise disjoint collection of sets
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Basic counting principles (cont.)
Example
By the Addition Principle, the number of odd numbers less than
10,000 is
|N1 | + |N2 | + |N3 | + |N4 |
Using the Multiplication Principle as we did in the previous example
We see that
|N1 | = 5, |N2 | = 45, |N3 | = 450
Thus the answer is 4500
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and
|N4 | = 4500
+ 450 + 45 + 5 = 5000
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Basic counting principles (cont.)
Example
How many
k -tuples
can be chosen from a set of
n
elements if
repetition is allowed?
k -tuples (s1 , s2 , . . . , sk ) in which each si comes
n elements and in which it's possible that si = sj
We seek the number of
from a xed set of
For each position in the
k -tuple,
we can choose any one of
n
dierent
elements
Hence, by the Multiplication Principle, there are
n| · n ·{z· · · · n} = nk
k
such
terms
k -tuples
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The Power Set
Denition
If
S
is a set, then we let 2
called the
power set
of
S
denote the set of all subsets of
S
S,
sometimes
Example
If
S = {a , b , c },
2
then
S = {∅, {a}, {b }, {c }, {a, b }, {a, c }, {b , c }, {a, b , c }}
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The Power Set (cont.)
The Multiplication Principle enables us to prove a formula for the
number of subsets of a set
Theorem
If S is a nite set, then
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S
2 = 2|S |
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The Power Set (cont.)
Proof.
Let
S = {x1 , x2 , . . . , x|S | }
With each subset
A⊆S
do the following:
Associate an |S |-tuple (a1 , a2 , . . . , a|S | ) with A
Where for i ∈ {1, 2, . . . , |S |} we set
(
ai =
0
1
if xi 6∈ A
if xi ∈ A.
Clearly, each subset corresponds to exactly one
|S |-tuple
|S |-tuple
and each
corresponds to one subset
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The Power Set (cont.)
Proof.
It follows that the number of subsets is equal to the number of such
|S |-tuples
Note each position in the
|S |-tuple
is chosen from a set
Thus we are asking the question How many
|S |-tuples
{0 , 1 }
of size 2
can be chosen
from a set of 2 elements?
|S |
From the second to last example, the answer is 2
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The Power Set (cont.)
Example
Let
S = {a, b, c }
The subset
{a , c }
The subset
{b }
and let
x1 = a, x2 = b,
x3 = c
and
corresponds to the 3-tuple
corresponds to the 3-tuple
(1, 0, 1)
(0, 1, 0)
According to previous theorem, the number of subsets of
S
is
S
2 = 2|S | = 23 = 8
This can be veried by counting the subsets
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The Pigeonhole Principle
The Pigeonhole Principle
We study a principle referred to as the
Pigeonhole Principle
Theorem
If n + 1 objects are placed into n boxes, then at least one box contains at
least two objects
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The Pigeonhole Principle
The Pigeonhole Principle (cont.)
Proof.
Suppose this is not the case
Then, each box contains at most one object
This implies that the number of objects is at most
n
This is a contradiction
Example
If we choose 13 people, then there are two who have their birthday in
the same month
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The Pigeonhole Principle
The Pigeonhole Principle (cont.)
Example
Suppose we have
n
How many of the 2
married couples
n
people must be selected to guarantee that we
have chosen at least one married couple?
Construct
n
boxes, each corresponding to one married couple
Each time we choose someone, place them into the box corresponding
to the couple they are a member of
The Pigeonhole Principle says that once we've chosen
n+1
people, at
least one box must contain 2 people
That is, we've chosen a married couple
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The Pigeonhole Principle
The Pigeonhole Principle (cont.)
Example
We choose 101 of the integers 1, 2, . . . , 200
We claim that among the integers chosen, there are two having the
property that one is divisible by the other
Each integer in
n
{1, 2, . . . , 200}
can be written in the form
n 2k ,
where
is an odd number between 1 and 199
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The Pigeonhole Principle
The Pigeonhole Principle (cont.)
Example
There are 100 odd integers in
{1, 2, . . . , 200}
So if we choose 101 numbers, by the Pigeonhole Principle, two of the
numbers we've chosen must have the same odd
n
That is, two of the numbers we've chosen are of the form
n2k2
We assume without loss of generality that
The number
n2k2
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divides the number
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n2k1
and
k1 ≥ k2
n2k1 ,
as desired
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The Pigeonhole Principle
The Strong Pigeonhole Principle
We now give a stronger form of the Pigeonhole Principle
Theorem
Let n1 , n2 , . . . , nk be positive integers
Let
n1 + n2 + · · · + nk − k + 1
objects are placed into k boxes
Then there is an integer i ∈ {1, 2, . . . , k } such that the ith box
contains at least ni objects
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The Pigeonhole Principle
The Strong Pigeonhole Principle (cont.)
Proof.
Assume, to the contrary, that this is not the case
Then for each
i ∈ {1, 2, . . . , k },
the
i th
box contains at most
ni − 1
objects
Thus the total number of objects is at most
(n1 − 1) + (n2 − 1) + · · · + (nk − 1) = n1 + n2 + · · · + nk − k
This is a contradiction
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The Pigeonhole Principle
The Strong Pigeonhole Principle (cont.)
Example
A basket of fruit is to be made up from apples, bananas, litchis, and
mangos
How many pieces of fruit must we place in the basket to be
guaranteed that there are at least three apples or at least two bananas
or at least ten litchis or at least ve mangos?
From the Strong Pigeonhole Principle, the answer is
3
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+ 2 + 10 + 5 − 4 + 1 = 17
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The Pigeonhole Principle
The Strong Pigeonhole Principle (cont.)
Example
Suppose that we choose
n2 + 1
integers from the integers 1, 2, . . . ,
n
Think of this in the following way:
We pick n2 + 1 integers
Each integer is put into a box marked from 1 to n depending on its
value
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The Pigeonhole Principle
The Strong Pigeonhole Principle (cont.)
Example
Note that
n2 + 1 = (n + 1) + (n + 1) + · · · + (n + 1) −n + 1
|
n
{z
}
terms
The Strong Pigeonhole Principle guarantees that at least one box
contains at least
n+1
objects
That is, at least one of the integers 1, 2, . . . ,
n
is chosen at least
n+1
times
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The Pigeonhole Principle
The Strong Pigeonhole Principle (cont.)
Let
a1 , a2 , . . . , ak
be a sequence of real numbers
Recall that a subsequence
ai1 , ai2 , . . . , ait , where
1
The sequence
of
a1 , a2 , . . . , ak
is of the form
≤ i1 < i2 < · · · < it ≤ k
a1 , a2 , . . . , ak
is
increasing if a1 ≤ a2 ≤ · · · ≤ ak and
decreasing if a1 ≥ a2 ≥ · · · ≥ at
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The Pigeonhole Principle
The Strong Pigeonhole Principle (cont.)
Example
Consider the sequence 8, 1, 3, 5, 9, 2, 6, 4, 7
Then 1, 5, 2, 7 is a subsequence, but 1, 2, 3, 4 is not
Note that 1, 3, 5, 9 is an increasing subsequence, and 8, 5, 2 is a
decreasing subsequence
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The Pigeonhole Principle
The Strong Pigeonhole Principle (cont.)
The Strong Pigeonhole Principle enables us to prove the following
interesting result, rst established by Erdös and Szekeres
Theorem
Every sequence a1 , a2 , . . . , an2 +1 of n2 + 1 real numbers contains an
increasing subsequence of length n + 1 or a decreasing subsequence of
length n + 1
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One-to-one functions and permutations
One-to-one functions
As mentioned before, a function is a relation in which each element of
the domain is related to exactly one element of the range
Let
f
be a function with the additional property that no two elements
of its domain are related to the same element of its range
That is,
x 6= y
We call
f
a
implies that
one-to-one
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f (x ) 6= f (y )
function, or an
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injection
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One-to-one functions and permutations
One-to-one functions (cont.)
If
f
is one-to-one, then the relation
f −1 = {(b, a) : (a, b) ∈ f }
is also a function
Conversely, if
If
f
f
is
not one-to-one, then f −
is an injection, then
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1
f −1
is called the
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is
not a function
inverse function
of
f
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One-to-one functions and permutations
One-to-one functions (cont.)
Example
Let
f1 = {(1, a), (2, c ), (3, b)}
Both
f1
and
f2
The function
and
f2 = {(1, a), (2, b), (3, b)}
are functions
f1
is an injection, while
This follows since the element
b
f2
is not
of the range of
f2
is related to both 2
and 3
The inverse of
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f1
is
f1−1 = {(a, 1), (c , 2), (b, 3))}
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One-to-one functions and permutations
One-to-one functions (cont.)
Example
The relation
f3 = {(x , x 2 ) : x ∈ R}
However, both
(2, 4)
and
(−2, 4)
is a function
are members of
f3 ,
so
f3
is not
one-to-one
The relation
f4 = {(x , 5x + 6) : x ∈ R}
Its inverse is the function
The relation
is a one-to-one function
f4−1 = {(x , (x − 6)/5 : x ∈ R}
f5 = {(x , tan(π(x − 1/2))) : x ∈ (0, 1)}
is a one-to-one
function
Its inverse is the function
f5−1 = {(x , (1/π) arctan x + 1/2) : x ∈ (−∞, ∞)}
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One-to-one functions and permutations
Permutations
Let
A
S
be a nite nonempty set
permutation
S
of
S
is a one-to-one function whose domain and range
are both
Example
Let
S = {1, 2, 3}
One permutation of
f (3) = 2
S
is the function
f
with
f (1) = 1, f (2) = 3,
and
This permutation can be denoted in the following manner
f =
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1
2
3
1
3
2
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One-to-one functions and permutations
Permutations (cont.)
The top row of the matrix is read as the domain and the bottom row
as the range
Thus, in general, this notation has this form
f =
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1
2
3
f (1) f (2) f (3)
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One-to-one functions and permutations
Permutations (cont.)
Example
There are six permutations of a set of three elements. The six
{1, 2, 3}
permutations of
f1 =
f4 =
1
2
3
1
2
3
1
2
3
1
3
2
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are:
f2 =
f5 =
1
2
3
2
3
1
1
2
3
3
2
1
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f3 =
f6 =
1
2
3
3
1
2
1
2
3
2
1
3
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One-to-one functions and permutations
Permutations (cont.)
Written this way, it's easy to nd the inverse of a permutation
To do so,
1
2
Interchange the top and bottom rows of the matrix
(If necessary) re-order by the elements in the (new) rst row
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One-to-one functions and permutations
Permutations (cont.)
Example
The inverse of
f2 =
is
1
2
3
2
3
1
2
3
1
1
2
3
After resorting the top row into ascending order, becomes the
permutation
In other words,
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1
2
1
3
1
2
= f3
f2−1 = f3
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One-to-one functions and permutations
Permutations (cont.)
Example
Similarly
f1−1 = f1
f3−1 = f2
f4−1 = f4
f5−1 = f5
f6−1 = f6
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One-to-one functions and permutations
Permutations (cont.)
f = f −1 ,
involution
If
as is the case with
f1 , f4 , f5 ,
and
f6 ,
then
f
is called an
Another way of writing permutations is as follows
For the permutation
2
3
4
2
4
1
3
we only write the bottom row,
Thus, we could write
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1
(2413)
f1 = (123), f6 = (213),
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etc.
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Counting permutations
Counting permutations
We now attempt to count permutations and similar mathematical
objects
Example
How many permutations of the set
S = {a, b, c }
are there?
As mentioned in the last example, we can think of each permutation
on
S
as a 3-tuple
(f (a), f (b), f (c ))
The question then becomes: How many such 3-tuples are there?
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Counting permutations
Counting permutations (cont.)
Example
The value
f (a )
may be any of three things:
Once we've assigned a value to
Since
f
is one-to-one,
f (b )
f (a ),
a, b ,
and
we must assign one to
cannot have the same value as
This means that there are only two possibilities for
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c
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f (b )
f (a )
f (b)
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Counting permutations
Counting permutations (cont.)
Example
Lastly, once we've chosen
neither
f (a)
nor
f (b )
f (b ),
we must specify
It follows that, whatever the values of
possible value for
f (c )
f (a )
and
f (c ),
f (b),
which can be
there's only one
From the Multiplication Principle, the total number of such 3-tuples is
3
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·2·1=6
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Counting permutations
Counting permutations (cont.)
Recall that
n factorial
is the function with rule 0!
=1
and, for
n ≥ 1,
n! = n · (n − 1) · (n − 2) · · · 3 · 2 · 1
So,
0!
=1
1!
=1
2!
=2
3!
=6
4!
= 24
5!
= 120
This gives us a compact way to write certain expressions down
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Counting permutations
Counting permutations (cont.)
n-set
An
is a set containing
n
elements
Example
How many permutations of an
n-set
Suppose that the elements of the
are there?
n-set
are
x1 , x2 , . . . , xn
As in the previous example, we can consider each permutation to be
an
n-tuple (f (x1 ), f (x2 ), . . . , f (xn ))
As before, we have
n
Once we've chosen
f (x1 ),
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choices for
f (x1 )
there are
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(n − 1)
possible choices for
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f (x2 )
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Counting permutations
Counting permutations (cont.)
Example
Once we've chosen
f (x2 ),
there are
(n − 2)
possible choices for
f (x3 ),
and so on
So we see that the number of permutations of an
n-set
is
n(n − 1)(n − 2) · · · 2 · 1 = n!
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Counting permutations
Counting permutations (cont.)
Example
= 720
permutations of the set
{a, b, c , d , e , f }
k -tuples
can be chosen from an
n-set
There are 6!
Example
How many
if repetition of
elements is not allowed?
If repetition of elements is allowed, we already know the answer to this
question is
nk
Suppose then that repetition is not allowed
We may choose any of
n
elements for the rst position in the
k -tuple
Once we've done that, we may choose any element for the second
position except the one we chose for the rst
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Counting permutations
Counting permutations (cont.)
Example
So for each choice for the rst position, there are
n−1
choices for the
second position
Once we've chosen the second position, there are
(n − 2 )
possibilities
for the third position, and so on
When we reach the
k th
position, there are
n−k +1
possibilities
From the Multiplication Principle, the number of such
k -tuples
is thus
n(n − 1)(n − 2) · · · (n − k + 2)(n − k + 1)
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