R. Johnsonbaugh

Discrete Mathematics 5th edition, 2001

Chapter 4Counting methods

and the pigeonhole principle

4.1 Basic principlesMultiplication 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

n1n2…nk

Addition principleLet 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 sets

k

X = Xj

j =1

has n1 + n2 + … + nk elements

4.2 Permutations and combinations

A permutation of n distinct elements x1, x2,…, xn is 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

r-permutationsAn r-permutation of n distinct elements is an

ordering 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 is

P(n,r) = n(n-1)(n-2)…(n-r+1)

Combinations

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

distinct elements An r-combination of X is an unordered

selection 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!

Catalan numbers

Eugene-Charles Catalan (1814-1894)

Catalan numbers are 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 10

Cn 1 1 2 5 14 42 132 429 1430 4862 16796

4.3 Algorithms for generating permutations and combinations

Lexicographic order: Given two strings = s1s2…sp and = t1t2…tq

Define < if p < q and si = ti for all i = 1, 2,…, p

Or for some i, si ti and for the smallest i, si < ti

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

then < and < .

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 possible outcomes

Probability

Probability of an event is the number of outcomes 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|

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.

Probability function A probability function P is a function from

the set of all outcomes (sample space S) to the interval [0, 1], in symbols

P : S [0, 1]

The probability of an event E S is the sum of the probabilities of every outcome in E

P(E) = P(x)

x E

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

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)

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)

Pattern recognition Pattern recognition places items into classes,

based on various features of 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.

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

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 Binomial coefficients and combinatorial identities

Theorem 4.7.1: Binomial theorem. For any real numbers 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

Theorem 4.7.6: For 1 < k < n, C(n+1,k) = C(n,k) + C(n,k-1)

Pascal’s 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

…

4.8 The pigeonhole principle First form: If k < n and n pigeons fly into k

pigeonholes, some pigeonhole contains at least two pigeons.

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) for some x1, x2 X, x1 x2.

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 values a1, a2,…, ak X such that f(a1) = f(a2) = … f(ak).

Example: n = 5, m = 3k = n/m = 5/3 = 2.