R. Johnsonbaugh Discrete Mathematics 5 th edition, 2001 Chapter 4 Counting methods and the pigeonhole principle

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

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

4.2 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

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 : 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 = {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!

Catalan numbers  Eugene-Charles Catalan ( )  Catalan numbers are defined by the formula C n = C(2n,n) / (n+1) for n = 0, 1, 2,… The first few terms are: n CnCn

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  < .

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 = {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

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 )

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)

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 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

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 !

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)

Pascal’s Triangle …

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(x 1 ) = f(x 2 ) for some x 1, x 2  X, x 1  x 2.

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.