ch4- Counting principle

7/29/2019 ch4- Counting principle

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R. JohnsonbaughDiscrete Mathematics

5th edition, 2001

Chapter 4

Counting methodsand the pigeonhole principle


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4.1 Basic principles

Multiplication principle

If an activity can be performed in k

successive steps,

Step 1 can be done in n1 ways

Step 2 can be done in n2 ways

Step k can be done in nk ways

Then: the number of different ways that the

activity can be performed is the product

n1n2nk


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Addition principle

Let X1, X2,, Xk be a collection of k

pairwise disjoint sets, each of which has nj

elements, 1 < j < k, then the union of

those setsk

X = Xjj =1

has n1 + n2+ + nk elements


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4.2 Permutations and combinations

A permutationof n distinct elements x1, x2,, xnis an ordering of the n elements. There are n!

permutations of n elements.

Example: there are 3! = 6 permutations of three

elements a, b, c:

abc bac cab

acb bca cba


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r-permutations

An r-permutation of n distinct elementsis anordering of an r-element subset of the n

elements x1, x2,, xn

Theorem 4.2.10:

For r < n the number of r-permutations of a set

with n distinct objects isP(n,r) = n(n-1)(n-2)(n-r+1) = n !/(n-r) !

If r = n, P(n,n) = n !


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Combinations

Let X = {x1, x2,, xn} be a set containing n

distinct elements

An r-combinationof X is an unorderedselection of r elements of X, for r < n

The number of r-combinations of X is the

binomial coefficient

C(n,r) = = n! / r!(n-r)! = P(n,r)/ r!


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Catalan numbers

Eugene-Charles Catalan (1814-1894)

Catalan numbersare defined by the formula

Cn

= C(2n,n) / (n+1)

for n = 0, 1, 2,

The first few terms are:

n 0 1 2 3 4 5 6 7 8 9 10Cn 1 1 2 5 14 42 132 429 1430 4862 16796


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4.3 Algorithms for generating

permutations and combinations

Lexicographic order:

Given two strings = s1s2sp and = t1t2tq Define < if

p < q and si = tifor all i = 1, 2,, p

Or for some i, si t

iand for the smallest i, s

i< t

i

Example: if = 1324, = 1332, = 132,

then < and < .


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4.4 Introduction to

discrete probability

An experiment is a process that yields an

outcome

An event is an outcome or a set of outcomes

from an experiment

The sample space is the event of all possibleoutcomes


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Probability

Probability of an eventis the number ofoutcomes in the event divided by the number of

outcomes in the sample space.

If S is a finite sample space and E is an event

(E is a subset of S) then the probability of E is

P(E) = |E| / |S|


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4.5 Discrete probability theory

When all outcomes are equally likely and

there are n possible outcomes, each one

has a probability 1/n.

BUT this is not always the case. When

all probabilities are not equal, then some

probability (possibly different numbers)must be assigned to each outcome.


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Probability function

A probability functionP is a function fromthe set of all outcomes (sample space S) tothe interval [0, 1], in symbols

P : S [0, 1]

The probability of an eventE S is the sumof the probabilities of every outcome in E

P(E) = P(x)x E


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Probability of an event

Given E S, we have

0 < P(E) < P(S) = 1

If S = {x1

, x2

,, xn

} is a sample space, then

n

P(S) = P(xi) = 1

i =1

If Ec is the complement of E in S, then

P(E) + P(Ec) = 1


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Events in a sample space

Given any two events E1 and E2 in a sample

space S. Then

P(E1 E2) = P(E1) + P(E2) P(E1E2)

We also have P() = 0

Events E1 and E2 are mutually exclusive if and

only if E1E2 = . In this case

P(E1E2) = P(E1) + P(E2)


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Conditional probability

Conditional probability is the probability of an

event E, given that another event F has

occurred, is called. In symbols P(E|F).

If P(F) > 0 then

P(E|F) = P(EF) / P(F)

Two events E and F are independent if

P(EF) = P(E)P(F)


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Pattern recognition Pattern recognitionplaces items into classes,

based on various featuresof the items.

Given a set of features F we can calculate the

probability of a class C, given F: P(C|F) Place the item into the most probable class, i.e.

the one C for which P(C|F) is the highest. Example: Wine can be classified as Table wine (T), Premium (R)

or Swill (S). Let F {acidity, body, color, price} Suppose a wine has feature F, and P(T|F) = 0.5, P(R|F) = 0.2

and P(S|F) = 0.3. Since P(T|F) is the highest number, this wine

will be classified as table wine.


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Bayes Theorem

Given pairwise disjoint classes C1, C2,,

Cn and a feature set F, then

P(Cj|F) = A / B, where

A = P(F|Cj)P(Cj)

n

and B = P(F|Ci)P(Ci)i = 1


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Generalized permutations

and combinations

Theorem 4.6.2: Suppose that a sequence of n

items has nj identical objects of type j,for 1< j < k. Then the number of orderings of S

is

____n!____

n1!n2!...nk!

4 7 Bi i l ffi i t


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4.7 Binomial coefficients

and combinatorial identities

Theorem 4.7.1: Binomial theorem. For any realnumbers a, b, and a nonnegative integer n:

(a+b)n = C(n,0)anb0 + C(n,1)an-1b1+

+ C(n,n-1)a1bn-1 + C(n,n)a0bn

n

(a+b)n = C(n,k) an-k bkk = 0

Theorem 4.7.6: For 1 < k < n,

C(n+1,k) = C(n,k) + C(n,k-1) andn

C(i,k) = C(n+1,k+1)i = k


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Pascals Triangle

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1


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4.8 The pigeonhole principle

First form: If k < n and n pigeons fly into kpigeonholes, some pigeonhole contains at least

two pigeons.


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Second form of the

pigeonhole principle

If X and Y are finite sets with |X| > |Y| and

f : X Y is a function, then f(x1) = f(x2) forsome x1, x2 X, x1 x2.


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Third form of the

pigeonhole principle

If X and Y are finite sets with |X| = n, |Y| = m

and k = n/m, then there are at least k valuesa1, a2,, ak X such that f(a1) = f(a2) = f(ak).

Example:

n = 5, m = 3

k = n/m = 5/3 = 2.

ch4- Counting principle

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