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Counting Chapter 4 The Basics of Counting 4.1 Sum Rule If a task can be done n1 ways Or the task can be done n2 ways there are n1 + n2 ways to do the task NOTE: Only one task is performed; it is performed via n1 OR n2 Sum Rule Examples You may pick one committee member The member may be from the faculty OR from the computer science student body There are 12 CS instructors There are 322 CS students There are 12 + 322 = 334 possible choices Sum Rule Examples You may pick a project from one of three lists: – 23 on the first – 15 on second – 19 on third There are 23 + 15 + 19 = 57 possible choices The sum rule is analogous to sequential for loops k=0; for (int i=0; i<n; i++) k = k +1; for (int i=0; i<m; i++) k = k +1; for (int i=0; i<p; i++) k = k +1; Final k value: k==n+m+p Product Rule A procedure can be broken down into 2 tasks The first task can be done n1 ways The second task can be done n2 ways there are n1 · n2 ways to do the procedure NOTE: both tasks are required to do the procedure n1 AND n2 Product Rule Examples A student brings 5 shirts and 4 pairs of pants to campus. How many outfits can he wear? 5 · 4 = 20 outfits Product Rule Examples A Chinese restaurant serves – 3 soups – 5 meat entrees – 6 vegetables A regular priced meal consists of one soup, one meat, and one vegetable. How many different regular meals could one order? There are 3 · 5 · 6 = 90 possible meals Product Rule Examples How many bit strings of length seven exist? _ _ _ _ _ _ _ 2 · 2 · 2 · 2 · 2 · 2 · 2 = 27 7 “holes” to be filled; 2 ways to fill each hole Product Rule Examples How many license plates are available if each plate consists of – 3 letters – 3 digits 26 · 26 · 26 · 10 · 10 · 10 17,576,000 The product rule is analogous to nested for loops k=0; for (int i=0; i<n; i++) for (int j=0; j<m; j++) for (int q=0; q<p; q++) k = k +1; Final k value: k==n·m·p Inclusion-Exclusion Principle To count the number of distinct elements in two sets – 1. 2. 3. where it it possible for some elements to be in both sets Add the number in the first set To the number in the second set Subtract number in both sets Same problem: count the ways of performing two tasks where performing the two tasks simultaneously is possible NOTE: The sum rule assumes sets are distinct (tasks may NOT be done at the simultaneously) Inclusion-Exclusion Example Of all CS students • 25 take C++ programming • 18 take discrete math • 6 take both How many CS students are there? • 25 + 18 is too many 25 + 18 - 6 = 37 CS students Inclusion-Exclusion |AB| = |A|+|B|- |AB| C++ 19 25 Discrete 6 12 18 Pigeonhole Principle 4.2 Pigeonhole Principle If k+1 or more objects are placed into k boxes – there is at least one box – containing two or more objects Pigeonhole Principle Examples In a group of 367 people – at least 2 people have the same birthday. (366 boxes, 367 objects) Given 27 English words – at least two begin with the same letter. (26 boxes, 27 objects) How many students must be in a class to – guarantee at least two score the same on a final – scored 0 to 100? 102 Generalized Pigeonhole Principle If n objects are placed in k boxes there is at least one box containing at least n/k objects Generalized Pigeonhole Examples A number 21 digits long – must have at least 3 digits which are the same n = 21 objects (digits) k = 10 boxes (0-9) n/k21/102.13 Of 100 people, at least 9 born same month n = 100 objects k = 12 boxes (months) n/k100/128.339 Generalized Pigeonhole Examples How many discrete students in a class of 11 are guaranteed to make the same grade? n = 11 objects k = 5 boxes n/k11/52.23 How many area codes guarantee a state with 25 million phones with have distinct # s? 720-7408; 10 million # s per area code n/k25M/10M2.53 Permutations and Combinations 4.3 Permutation An ordered arrangement of a set of distinct objects. List the permutations of {R,G,B} – {B,G,R} – {G,B,R} – {R,B,G} {B,R,G} {G,R,B} {R,G,B} r-Permutation An ordered arrangement of r elements of a set – – – – Select r objects from a set of n elements Do not select the same object twice Permute the r objects ORDER MATTERS Perform a 2-permutation on {R,G,B} {B,G} {B,R} {G,R} {G,B} {R,B} {R,G} Number of r-Permutations The number of r-Permutations in a set with n elements: P(n,r) = n(n-1)(n-2) … (n - r + 1) = n! (n r )! Note: P(n,n) = n! P(n,1) = n Permutation Examples How many possible arrangements exist of the letters: ABCEILNRSTU? P(n,n) = P(11,11) = n! = 11! = 11 10 9 8 7 6 5 4 3 2 1 3,628,800 Find the word, get three bonus points. (16/class) Permutation Examples 10 students compete for gold, silver, and bronze programming medals. How many different ways might the programmers finish? P(n,r) = P(10,3) = 10! / (10-3)! = 10 9 8720 Permutation Examples How many ‘words’ of 4 letters can be obtained from the letters: NORTHEAS P(n,r) = P(8,4) = 8! / (8-4)! = 8 7 6 5 1680 r-Combination A selection of r elements of a set – – – – Select r objects from a set of n elements Do not select the same object twice Equate lists with the same elements: {B,R}={R,B} ORDER DOES NOT MATTER List all 2-combinations on {R,G,B} – {B,G} – {B,R} – {G,R} Number of r-Combinations The number of r-Combinations in a set with n elements: n! C(n,r) = (n r )! r ! Note: C(n,n) = C(n,0) = 1 C(n,1) = C(n, n-1) = n Combination Examples What 3-combinations exist of the set {1,2,3,4}? {2,3,4} {1,3,4} {1,2,4} {1,2,3} Combination Examples What 2-combinations exist of the set {1,2,3,4}? {1,2} {1,3} {1,4} {2,3} {2,4} {3,4} Combination Examples A Chinese restaurant allows one to order 2 of 8 main dishes. How many different combinations of 2 dishes are available? C(n,r) = n! / ((n-r)! r!) = 8! / ((8-2)! 2!) = 8 7/2 28 Combination Examples How many programming teams (4 members) can be selected from 10 CS students? C(10,4) = 10! / ((10-4)! 4!) = (10 9 8 7)/(4 3 2) 210 Combinations Example How many poker hands are there? C(52,5) = 52! / ((52-5)! 5!) = (52 51 50 49 48) /(5 43 2) 2,598,960 What is the Probability of drawing a flush in a poker game? How many diamond flush hands? C(13,5) = 13! / ( (13-5)! 5! ) = (13 12 11 10 9) /(5 4 3 2) 1287 Hands of any flush? |E| = 4 12875148 |S| = C(52,5) = 2,598,960 p(E) = |E| / |S| = 5,148 / 2,598,960 = 0.0019807 = 0.2% or 2 out of 1000 Binomial Coefficients 4.4 Binomial Coefficients C(n, r) “n CHOOSE r” n r is called a binomial coefficient The value of C(n, r) occurs as coefficients in the expansion of powers of binomial expressions such as (a + b)n Pascal’s Triangle Binomial Theorem Given variables, x and y and positive integer n n ( x y ) C ( n, j ) x n n j y j j 0 n n n n 1 n n 2 2 n n 1 n n x x y x y ... xy y 0 1 2 n 1 n Binomial Examples Expand (x+y)4. 4 ( x y ) C ( 4, j ) x 4 4 j y j j 0 4 4 4 3 4 2 2 4 4 4 3 x x y x y x y y 0 1 2 3 4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4 Binomial Examples What is the coefficient of x12y13 in the expansion of (x+y)25? C(25,13) = 25! / (13! 12!) = 5,200,300 Binomial Examples What is the coefficient of x12y13 in the expansion of (2x-3y)25? 25 (2 x 3 y ) C(25, j ) (2 x ) 25 25 j j 0 C(25,13) 212 (-3)13 = - ( 25! 212 313) / (13! 12!) ( 3 y ) j Discrete Probability 5.1 Finite Probability Experiment – a procedure that yields one of a given set of possible outcomes Sample Space – the set of possible outcomes of an experiment Event – a subset of the sample space Probability of event E The probability of event E – a subset of sample space S – of equally likely outcomes p(E) = |E| / |S| Probability Example An urn contains 4 blue and five red legos. What is probability a blue lego is chosen? S is the set of all possible outcomes |S| = 9 E is a subset of S which results in blue choices |E| = 4 p(E) = |E| / |S| = 4/9 Probability Example What is the probability that when two dice are rolled, the sum of the numbers is 7? S is the set of all possible outcomes |S| = 6 636 E is a subset of S which results in total of 7 (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) |E| = 6 p(E) = |E| / |S| = 6 / 36 = 1/6 Probability Example How many poker hands are there? C(52,5) = 52! / ((52-5)! 5!) = (52 51 50 49 48) /(5 43 2) 2,598,960 What is the Probability of drawing a flush in a poker game? How many diamond flush hands? C(13,5) = 13!/((13-5)! 5!) = (13 12 11 10 9) /(5 4 3 2) 1287 Hands of any flush? |E| = 4 12875148 |S| = C(52,5) = 2,598,960 p(E) = |E| / |S| = 5,148 / 2,598,960 = 0.0019807 = 0.2% or 2 out of 1000 Lottery Example Win a big prize if you pick a correct 4 digit #. Win a smaller prize if you pick any 3 of the 4 digits corrects. S is the set of all possible outcomes |S| = 9999 + 1 The winner of the big prize must choose all 4 correctly |E| = 1 p(E) = |E| / |S| = 1/10000 Lottery Example Win a smaller prize if you pick any 3 of the 4 digits corrects. The winner of the smaller prize must choose 3 correctly 1 1 1 9 (Rule out correct # in last digit) 1191 1911 9111 p(E) = |E| / |S| = 9 4/10000 = .36 % Lotto Example Choose 6 numbers between 1 and 40 (1) Winner picks all 6 correctly C(40, 6) = 40! / (34! 6!) = 3,838,380 Probability 1/ 3,838,380 = 0.000026% Complementary Event E Given E is an event in a sample space S The probability of E, – the complementary event of E p(E) = 1 - p(E) NOTE: E E = U Complementary Event Example 10 bits are generated randomly What is probability that one of these bits is 0? How many bit strings DO NOT have a zero? 1 How many bit strings of length 10 exist? 1024 Let E be event that strings DO NOT have zero p(E) = 1 / 1024 Let E be event that strings have a zero p(E) = 1 - p(E) = 1 - (1 / 1024) = 1023 / 1024 = 99.9% Probability of Two Events Let E1 and E2 be events in sample space S. p(E1 E2) = p(E1) + p(E2) - p(E1 E2) Two Events Probability Example Pick a number (randomly) between 1-100 What is probability number is divisible by either 2 or 5? p(E1) = 50 / 100 p(E2) = 20 / 100 Numbers divisible by 10 are divisible by 2 and 5 p(E1 E2) = 10 / 100 p(E1 E2) = p(E1) + p(E2) - p(E1 E2) = (50 + 20 - 10) / 100 = 60 / 100 = 3 / 5