Download presentation
Presentation is loading. Please wait.
Published byAllison Osborne Modified over 9 years ago
1
R. Johnsonbaugh Discrete Mathematics 5 th edition, 2001 Chapter 4 Counting methods and the pigeonhole principle
2
4.1 Basic principles Multiplication principle If an activity can be performed in k successive steps, Step 1 can be done in n 1 ways Step 2 can be done in n 2 ways … Step k can be done in n k ways Then: the number of different ways that the activity can be performed is the product n 1 n 2 …n k
3
Addition principle Let X 1, X 2,…, X k be a collection of k pairwise disjoint sets, each of which has n j elements, 1 < j < k, then the union of those sets k X = X j j =1 has n 1 + n 2 + … + n k elements
4
4.2 Permutations and combinations A permutation of n distinct elements x 1, x 2,…, x n 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: abcbaccab acbbcacba
5
r-permutations An r-permutation of n distinct elements is an ordering of an r-element subset of the n elements x 1, x 2,…, x n 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)
6
Combinations Let X = {x 1, x 2,…, x n } 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!
7
Catalan numbers Eugene-Charles Catalan (1814-1894) Catalan numbers are defined by the formula C n = C(2n,n) / (n+1) for n = 0, 1, 2,… The first few terms are: n012345678910 CnCn 112514421324291430486216796
8
4.3 Algorithms for generating permutations and combinations Lexicographic order: Given two strings = s 1 s 2 …s p and = t 1 t 2 …t q Define < if p < q and s i = t i for all i = 1, 2,…, p Or for some i, s i t i and for the smallest i, s i < t i Example: if = 1324, = 1332, = 132, then < and < .
9
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
10
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|
11
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.
12
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
13
Probability of an event Given E S, we have 0 < P(E) < P(S) = 1 If S = {x 1, x 2,…, x n } is a sample space, then n P(S) = P(x i ) = 1 i =1 If E c is the complement of E in S, then P(E) + P(E c ) = 1
14
Events in a sample space Given any two events E 1 and E 2 in a sample space S. Then P(E 1 E 2 ) = P(E 1 ) + P(E 2 ) – P(E 1 E 2 ) We also have P( ) = 0 Events E 1 and E 2 are mutually exclusive if and only if E 1 E 2 = . In this case P(E 1 E 2 ) = P(E 1 ) + P(E 2 )
15
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(E F) / P(F) Two events E and F are independent if P(E F) = P(E)P(F)
16
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.
17
Bayes’ Theorem Given pairwise disjoint classes C 1, C 2,…, C n and a feature set F, then P(C j |F) = A / B, where A = P(F|C j )P(C j ) n and B = P(F|C i )P(C i ) i = 1
18
Generalized permutations and combinations Theorem 4.6.2: Suppose that a sequence of n items has n j identical objects of type j, for 1< j < k. Then the number of orderings of S is ____n!____ n 1 !n 2 !...n k !
19
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)a n b 0 + C(n,1)a n-1 b 1 + … + C(n,n-1)a 1 b n-1 + C(n,n)a 0 b n Theorem 4.7.6: For 1 < k < n, C(n+1,k) = C(n,k) + C(n,k-1)
20
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 …
21
4.8 The pigeonhole principle First form: If k < n and n pigeons fly into k pigeonholes, some pigeonhole contains at least two pigeons.
22
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(x 1 ) = f(x 2 ) for some x 1, x 2 X, x 1 x 2.
23
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 a 1, a 2,…, a k X such that f(a 1 ) = f(a 2 ) = … f(a k ). Example: n = 5, m = 3 k = n/m = 5/3 = 2.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.