MAT3707. Tutorial letter 202/1/2017 DISCRETE MATHEMATICS: COMBINATORICS. Semester 1. Department of Mathematical Sciences MAT3707/202/1/2017
|
|
- Damian Blake
- 5 years ago
- Views:
Transcription
1 MAT3707/0//07 Tutorial letter 0//07 DISCRETE MATHEMATICS: COMBINATORICS MAT3707 Semester Department of Mathematical Sciences SOLUTIONS TO ASSIGNMENT 0 BARCODE Define tomorrow university of south africa
2 SOLUTIONS TO ASSIGNMENT 0 (SEMESTER ) CLOSING DATE: April 07 UNIQUE NR: 8676 The answers to only some of the questions will be marked How many different -letter words (sequences) are there with no repeated letters, formed from the 6-letter alphabet? (3) This is just the number of ways of choosing objects from 6 different objects where order matters which is P (6, ) (a) How many ways are there to seat six different boys and six different girls along one side of a long table with seats? () This is the same as the number of permutations of distinct objects:! (b) How many ways if the boys and girls alternate? () There are two types of seating arrangements: BGBGBGBGBGBG and GBGBGBGBGBGB For each of these types of arrangements there are 6! ways of arranging the boys, and 6! ways for arranging the girls By the Multiplication Rule there are then 6! 6! (6!) ways of making an arrangement in each of the cases Thus in total there are (6!) + (6!) (6!) ways 3 How many ways are there to pick a subset of different letters from the 6-letter alphabet? (3) This is just the number of ways of choosing objects from 6 different objects where order does not 6 matter, which is There are white balls and 6 red balls in a container (a) How many different ways are there to select a subset of 6 balls, assuming the 8 balls are all different? (3) Order does not matter, so the answer is colours are irrelevant) 8 (Note that since the balls are all different, their 6
3 MAT3707/0//07 (b) What is the probability that the selection has whites and reds? (3) 6 There are ways of choosing the white balls and ways of choosing the red balls By 6 the Multiplication Principle there are ways of choosing Thus the probability is )( 6 ) ( ( 8 6 ) 5 There are women and 6 men who will split up into five-person teams How many ways are there to do this so that there is (at least) one women on each team? () Note that it is sufficient to choose the first team, since the second team is then fixed There are exactly three possible disjoint cases to consider: Case A woman and men on the first team Case B women and 3 men on the first team Case C 3 women and men on the first team (Note that women on the first team would imply no women on the second team, which is not allowed) 6 In Case A there are ways of choosing the woman and ways of choosing the men By 6 the Multiplication Principle there are ways in case A 6 Similarly, in Case B there are ways 3 6 And in Case C there are ways By the Addition Principle there are + + ways in total (a) What is the probability that a five-card hand has at least one card of each suit? (3) A five-card hand with at least one card of each suit, must have two cards of one suit, and one card of each of the three remaining suits, ie it must be one of,,, 3 For each of these four cases, we choose the two cards of the repeating suit in ways (order 3 does not matter in a hand), and then each of the remaining three cards in ways (recall that there are 3 cards of each suit) By the Multiplication Principle there are ways of choosing a hand with at least one cared of each suit 3
4 In total there are 5 ways of choosing a 5-card hand (order does not matter) Thus the proba- 5 bility is ) 3 ( 3 3 ( 5 ) 5 (b) What is the probability that a six-card hand has at least one card of each suit? (5) The possibilities for a 6-card hand with at least one card of each suit is somewhat more complicated There are two main possibilities: Case A Three cards of one suit and one card of each of the remaining three suits (this is four possibilities, one for each of the suits that occurs three times, eg ), Case B Two cards of one suit, two cards of another suit, and one card of each of the remaining two suits ( possibilities, one for each pair of suits occurring twice, eg ) In Case A we build a 6-card hand as follows: Choose which suit occurs three times: possibilities (as already mentioned) 3 Choose the 3 cards for the triply occurring suit: Choose a card for each of the remaining three suits: 3 ( 3 )( 3 )( The Multiplication Principle gives 3 In Case B we build a 6-card hand as follows: Choose which two suits occurs twice: possibilities (as already mentioned) 3 3 For each of these two suits, choose the two cards: 3 3 Choose a card for each of the remaining two suits: 3 3 The Multiplication Principle gives The Addition Principle then gives (adding up Cases A and B) that there are in total ( )( 3 ) ( 3 six-card hands with at least one card of each suit Thus the probability is ( 3 3 ) 3 ( ) ( 3 ) )( ( 5 ) 6 ) )
5 MAT3707/0//07 7 What fraction of all arrangements of GRACEFUL have (a) F and G appearing side by side? () We combine F and G as one letter and call it X Now X together with R, A, C, E, U and L are 7 distinct letters which has can be arranged in 7! ways But X can be FG or GF which gives 7! different arrangements The total number of arrangements of the 8 distinct letters of GRACEFUL is 8!, so the fraction in which F and G appears side by side is 7! 8! (b) no pair of consecutive vowels? (5) First we arrange the five consonants G, R, C, F, L in 5! ways Think of them now as C, C, C 3, C, C 5 Since the vowels are not allowed to appear consecutively we can think of the consonants as separating the vowels So there are six possible places where we can put the vowels: before, in between and after the five consonants: C C C 3 C C 5 6 So first we pick three of these six possible places: ways 3 Then we arrange the three vowels: 3! ways 6 So by the Multiplication Principle there are 5! 3! ways in which to do this 3 The fraction of arrangements with no pair of consecutive vowels therefore is 8 Find the number of integer solutions to the inequality 5! ( 6 3) 3! 8! x + y + z 5 under the constraints x, y, z 5 (5) An inequality like this can always be converted into an equation by introducing an extra variable t 5 x y z Thus we have that x + y + z + t 5, where t is also an integer and it follows from the given inequality that t 0 Thus the constrains to this equation are x, y, z 5, t 0 We can model this situation as placing 5 identical objects into distinct containers, with at least 5 objects in each of the first three containers First place 5 objects into each of( the first three ) containers Then objects still have to be placed This gives 0 0 5
6 9 How many ways are there to pick a group of n people from 00 people (each of a different height) and then pick a second group of m other people such that all people in the first group are taller that the people in the second group? (6) Since the people have different heights, they are different objects, so they can be numbered from short to tall by {,, 3,, 00} We now have to find the number of ways of choosing a subset of n numbers and a second subset of m numbers from this set such that the numbers in the first subset are all smaller than the numbers in the second subset Note the following important observation: If we choose a subset of n + m numbers, then the n smallest numbers are already fixed, as are the largest m Thus choosing the two subsets ( is) exactly the same as choosing a single subset of n + m numbers 00 Thus the number of ways is n + m 0 How many ways are there to invite one of three different friends over for dinner on six successive nights such that no friend is invited more than three times? (6) If we let the 6 nights be 6 objects and the friends three types,,3 with a night being of type i if friend i is invited over for that night, then a way of inviting the 3 friends for the 6 nights is the same as the number of arrangements of these 6 objects, with say r of type, r of type, and r 3 of type 3 So we are counting arrangements with repetition The only problem is that r, r, r 3 do not have fixed values, since nothing specifies the number of nights r i on which friend i is invited So we just list all the possible values for the r i, keeping in mind that r + r + r 3 6 First note that if we don t take order into account, there are essentially only three solutions to r + r + r 3 6, namely , , + + 6, and by Theorem on page 96 of the Tucker there are A 6! ways of arranging 6 objects, 0 of type, 3 of type and 3 of type 3, 0!3!3! 6! B ways of arranging 6 objects, of type, of type and 3 of type 3,!!3! 6! C ways of arranging 6 objects, of type, of type and of type 3!!! But since the three terms on the left-hand side can be permuted, we really have 3 solutions as in Case A (one for each of , , ), 3! solutions as in Case B (one for each permutation of,, 3), and just solution as in Case C (there is just one way of writing + + ) Thus the answer is 3 6! 0!3!3! + 3! 6!!!3! + 6!!!! 6
7 MAT3707/0//07 How many arrangements of six 0 s, five s and four s are there in which (a) the first 0 precedes the first? () 5 First we position the four s among the 5 positions: ways Then we put a 0 in the first of the remaining positions: way 0 Finally we pick five other positions for the remaining 0 s: ways 5 Thus the answer is ( 5 )( 0 5 (b) the first 0 precedes the first, which precedes the first? (6) Put a 0 in the first position: way Next we pick five other positions for the remaining 0 s: ways 5 Now we put a in the first of the remaining positions: way And finally we pick four other positions for the remaining s: Thus there are 8 5 ways to do this ) 8 ways How many ways are there to distribute 5 (identical) apples, 6 (identical) oranges and 7 (identical) pears amongst people (a) without restriction? (3) We can distribute 5 apples among persons in ways, 6 oranges in ways and 7 pears in ways By the multiplication Principle there are ways to distribute them together (b) with each person getting at least one pear? () We first give each of the persons a pear (so that each one can have at least one pear; this can be done in one way), and then distribute the 5 apples, 6 oranges and the remaining 7 3 pears as in the previous question to obtain
8 3 Build a generating function for a r, the number of distributions of r identical objects into (a) five different boxes with at most objects in each box (3) Since the objects are identical, we use ordinary generating functions We model the problem as the number of integer solutions to e + e + e 3 + e + e 5 r, 0 e i Here e i is the number of objects going into box i We want to construct a product of polynomial factors such that when multiplied out, we obtain all the products x e x e x e 3 x e x e 5 with each e i satisfying the inequalities 0 e i Since such a product has five factors, we need five polynomial factors Each such polynomial needs to have all the x e i with e i in the admissable range of 0,,, 3, ie each polynomial must be x 0 + x + x + x 3 + x Thus the generating functions is (x 0 + x + x + x 3 + x ) 5 (b) four different boxes with between 3 and 8 objects in each box (3) We want the know the number of integer solutions to e + e + e 3 + e r, 3 e i 8 Now we need four polynomial factors, each of which is of the form x 3 + x + x 5 + x 6 + x 7 + x 8 Thus the generating function is (x 3 + x + x 5 + x 6 + x 7 + x 8 ) (c) seven different boxes with at least object in each box (3) We want the know the number of integer solutions to e + e + e 3 + e + e 5 + e 6 + e 7 r, e i Thus we need seven factors Note that now e i has no upper bound: it can be any one of,, 3, Thus each factor is strictly speaking not a polynomial anymore, but rather a power series, as there are infinitely many terms: x + x + x 3 + x + Thus the generating function is (x + x + x 3 + x + ) 7 (d) three different boxes with at most 5 objects in the first box (3) We want the know the number of integer solutions to e + e + e 3 r, e i 0, e 5 Thus we need three factors For the first factor we see that the possible values of e are 0,,, 3,, 5, thus the factor is x 0 + x + x + x 3 + x + x 5 For the other two factors e i has no restriction except e i 0, thus both are x 0 + x + x + x 3 + Thus the generating function is (x 0 + x + x + x 3 + x + x 5 )(x 0 + x + x + x 3 + ) 8
9 MAT3707/0//07 Use generating functions to find the number of ways to distribute r jelly beans among eight children if (a) each child gets at least one jelly bean (5) We model it as an equation e + e + + e 8 r, e i Thus there are 8 factors in the generating function, each equal to x + x + x 3 + Thus the generating function is (x + x + x 3 + ) 8 x 8 ( + x + x + ) 8 x 8 ( x) 8 and we want the coefficient of x r This is the same as the coefficient of x r 8 in is r r 8 r r 8 ( x) 8, which (b) each child gets an even number of beans (5) We still have the above equation e + e + + e 8 r, but now the constraints are that each e i must be even Thus each of the 8 factors now must be + x + x + x 6 + Thus the generating function is ( + x + x + x 6 + ) 8 8 x ( x ) ( 8 ) s x + x + + x s + s and we want the coefficient of x r If r is odd, then there is no power x r occurring, thus the coefficient is 0 Thus the number of ways of distributing an odd number of r jelly beans with each child getting an even number is 0, which agrees with common sense ( s + 8 If r is even, then r s for some s, and then the coefficient is r + 7 ) s r Thus the number of ways of distributing an even number of r jelly beans with each child getting an even number is ( r Use generating functions to find the number of ways there are to distribute 8 different toys among children if the first child gets at least toys (6) r ) 9
10 The toys are different, so we may model each distribution as an arrangement of length 8 of objects, with the first object occurring at least twice The exponential generating function of the number of arrangements of length r of objects, with the first object occurring at least twice, is ( x! + x3 3! + ) ( + x + x! + x3 3! + (e x x)(e x ) 3 (e x x)e 3x e x e 3x xe 3x and we determine the coefficient of x r /r! term by term The coefficient of x r /r! in ) 3 e x + x + (x)! + (x)3 3! + + (x)r r! + is r The coefficient of x r /r! in is 3 r The coefficient of x r /r! in e 3x + 3x + (3x)! xe 3x x ( + 3x + (3x)! x + 3x + 3 x 3! + (3x)3 3! + (3x)3 3! + 33 x! + + (3x)r r! + + (3x)r r! r x r+ + r! ) + is not so simple to see, so we first find the coefficient of x r+ This is 3r Thus the coefficient of r! xr is 3 r (we just replace r + by r) Finally, to find the coefficient of (r )! xr /r!, we have to multiply by r! and we find 3r r! (r )! 3r r Adding the coefficients up we obtain that the coefficient of x r /r! in e x e 3x xe 3x is r 3 r 3 r r Don t stop here: this is not the answer! Recall that the original question asked for 8 toys, not r toys So we substitute 8 into r and the answer is (a) Find a recurrence relation for the number of ways the elf in Example in Section 7 of Tucker can climb n stairs if each step covers either or or 3 stairs () 0
11 MAT3707/0//07 Let the number of ways to climb n stairs be a n Thus a (only one way), a (either + or ) and a 3 ( + + or + or + or 3) We now find the recurrence relation Consider the last step taken The number of ways n stairs can be climbed if the last step covers stair is exactly the number of ways n stairs can be climbed (just ignore the last step), which is a n Similarly, the number of ways n stairs can be climbed if the last step covers stairs is exactly the number of ways n stairs can be climbed, which is a n And similarly, the number of ways n stairs can be climbed if the last step covers 3 stairs is exactly the number of ways n 3 stairs can be climbed, which is a n 3 These three cases are disjoint and cover all possibilities (the last step must cover either, or 3 stairs) Thus by the addition principle, the total number of ways to climb n stairs is a n +a n + a n 3 Thus the recurrence relation is a n a n + a n + a n 3 (b) How many ways are there for the elf to climb 5 stairs? () We now use the recurrence relation and the first few values that we calculated in (a) a 5 a + a 3 + a (by the recurrence relation applied to n 5) (a 3 + a + a ) + a 3 + a (applied to n ) a 3 + a + a (by values found in (a)) 7 (a) Find a recurrence relation (with initial conditions) for the number of n-digit binary sequences (sequences of {0, }) with no pair of consecutive s (6) Call the number of such n-digit sequences a n There are two disjoint cases: either such an n-digit sequence begins with a 0 or it begins with a If it begins with a 0, then the remaining (n )-digit sequence following the 0 is just an arbitrary (n )-digit sequence with no pair of consecutive s Thus there are a n such sequences If it begins with a, we have to be careful: since the first digit is a and consecutive s are not allowed, the second digit is forced to be 0 Thus we have a sequence starting with 0 Then, exactly as the case of a sequence starting with 0, the remaining (n )-digit sequence is an arbitrary sequence of length n with no pair of consecutive s Thus there are a n such sequences The addition principle then gives that the recurrence relation is a n a n + a n, n 3 The first two values that are not covered by the recurrence relation is a (the two -digit sequences obviously have no pair of consecutive s) and a 3 (of the four -digit sequences only has a pair of consecutive s)
12 (b) Find a recurrence relation (with initial conditions) for the number of n-digit ternary sequences (sequences of {0,, }) with no pair of consecutive s or consecutive s (6) Call the number of such n-digit sequences a n Not allowing consecutive s or s implies that if you take such an n -digit sequence and if it ends in a 0, then you can add any one of 0, and to it, but if it ends in a, you can only add 0 or to it, and similarly, if it ends in a, you can only add 0 or to it So the only possibility where the last two digits are the same, is when they are 00 Thus the first n digits can be arbitrary, as long as no pair of consecutive s or s occur Thus there are a n such sequences In the remaining cases, the last two digits of the n-digit sequences are not the same This means that if you take any arbitrary n -digit sequence with no pair of consecutive s or s, you can add one of two different digits to the end of it (If it ends in a 0 you can add or to it, if it ends in you can add 0 or to it and if it ends in a you can add 0 or to it) Thus there are a n such sequences Thus in total there are a n + a n ternary sequences without consecutive s or s Thus the recurrence relation is a n a n + a n You can also calculate the first two values not covered by the recurrence relation: a 3 (the three -digit sequences obviously have no pair of consecutive s or s) and a 7 (of the nine -digit sequences only the sequences and have a pair of consecutive s or s) 8 Solve the recurrence relation a n 3a n + n, a (5) This is an inhomogeneous linear recurrence relation The corresponding homogeneous recurrence relation is a n 3a n with general solution a n A3 n We have f(n) n, thus a particular solution to the inhomogeneous relation is A n B n + B 0 Substitute into the recurrence relation: B n + B 0 3(B (n ) + B 0 ) + (n ) 3(B n B + B 0 ) + n (3B )n 3B + 3B 0 + n The equation is true for all values of n On both sides we have a polynomial Thus the coefficients on n must be equal: B 3B + and the constant coefficients must be equal: B 0 3B + 3B 0 From the first equation we obtain B From the second equation we then have B 0 3( ) + 3B 0 which gives B 0
13 MAT3707/0//07 Thus A n n Thus the general solution is a n A3 n n We determine A from the initial value a : A3, thus A 7 6 The solution is therefore a n 7 6 3n n 9 Use inclusion-exclusion to determine the number of 9-digit sequences of the digits,, 3, with each of the digits occurring at least once (6) Let A be the set of 9-digit sequences not using, A the set not using, A 3 the set not using 3 and A the set not using We want N(A A A 3 A ), ie the number of 9-digit sequences of,, 3, with each digit occurring at least once Then N 9 (all sequences) N(A A A 3 A ) N N(A ) N(A ) N(A 3 ) N(A ) +N(A A ) + N(A A 3 ) + N(A A ) +N(A A 3 ) + N(A A ) + N(A 3 A ) N(A A A 3 ) N(A A A ) N(A A 3 A ) N(A A 3 A ) +N(A A A 3 A ) N(A ) N(A ) N(A 3 ) N(A ) 3 9 (sequences using 3 digits) N(A A ) N(A 3 A ) 9 (sequences using digits) N(A A A 3 ) N(A A 3 A ) (only one sequence using only one digit) N(A A A 3 A ) 0 (no sequence using no digits) Hence N(A A A 3 A ) Use inclusion-exclusion to determine how many ways there are to choose 8 balloons from a collection of blue, red and green balloons (that are identical apart from their colour) if there must be at most seven balloons of each colour (6) Call the colours,, 3, and let A i be the set of selections with more than 7 of colour i (ie at least 8 balls of colour i) Now 8 + N 8 (total number of selections without any restriction), 8 3
14 and for each i,, 3,, 0 + N(A i ) 0 Then 3 (number of selections with at least 8 of colour i) 0 N(A A A 3 A ) N N(A ) N(A ) N(A 3 ) N(A ) + + For any two i, j, N(A i A j ) colour i and at least 8 of colour j) N(A A ) + N(A A 3 ) + N(A A ) +N(A A 3 ) + N(A A ) + N(A 3 A ) N(A A A 3 ) N(A A A ) N(A A 3 A ) N(A A 3 A ) +N(A A A 3 A ) 5 ( the number of selections with at least 8 of For any three i, j, k, N(A i A j A k ) 0 ( it is impossible to choose only 8 with at least 8 of each of the colours i, j, k) Similarly N(A A A 3 A ) 0 Hence N(A A A 3 A )
Mat 344F challenge set #2 Solutions
Mat 344F challenge set #2 Solutions. Put two balls into box, one ball into box 2 and three balls into box 3. The remaining 4 balls can now be distributed in any way among the three remaining boxes. This
More informationCONTENTS CONTENTS PAGES 11.0 CONCEPT MAP A. PERMUTATIONS a EXERCISE A B. COMBINATIONS a EXERCISE B PAST YEAR SPM
PROGRAM DIDIK CEMERLANG AKADEMIK SPM ADDITIONAL MATHEMATICS FORM 5 MODULE 11 PERMUTATIONS AND COMBINATIONS 0 CONTENTS CONTENTS PAGES 11.0 CONCEPT MAP 2 11.1 A. PERMUTATIONS 3 11.1a EXERCISE A.1 3 11.2
More informationCS100: DISCRETE STRUCTURES. Lecture 8 Counting - CH6
CS100: DISCRETE STRUCTURES Lecture 8 Counting - CH6 Lecture Overview 2 6.1 The Basics of Counting: THE PRODUCT RULE THE SUM RULE THE SUBTRACTION RULE THE DIVISION RULE 6.2 The Pigeonhole Principle. 6.3
More informationLEVEL I. 3. In how many ways 4 identical white balls and 6 identical black balls be arranged in a row so that no two white balls are together?
LEVEL I 1. Three numbers are chosen from 1,, 3..., n. In how many ways can the numbers be chosen such that either maximum of these numbers is s or minimum of these numbers is r (r < s)?. Six candidates
More informationCSCI 2200 Foundations of Computer Science (FoCS) Solutions for Homework 7
CSCI 00 Foundations of Computer Science (FoCS) Solutions for Homework 7 Homework Problems. [0 POINTS] Problem.4(e)-(f) [or F7 Problem.7(e)-(f)]: In each case, count. (e) The number of orders in which a
More informationExercises Exercises. 1. List all the permutations of {a, b, c}. 2. How many different permutations are there of the set {a, b, c, d, e, f, g}?
Exercises Exercises 1. List all the permutations of {a, b, c}. 2. How many different permutations are there of the set {a, b, c, d, e, f, g}? 3. How many permutations of {a, b, c, d, e, f, g} end with
More informationCreated by T. Madas COMBINATORICS. Created by T. Madas
COMBINATORICS COMBINATIONS Question 1 (**) The Oakwood Jogging Club consists of 7 men and 6 women who go for a 5 mile run every Thursday. It is decided that a team of 8 runners would be picked at random
More informationMA 524 Midterm Solutions October 16, 2018
MA 524 Midterm Solutions October 16, 2018 1. (a) Let a n be the number of ordered tuples (a, b, c, d) of integers satisfying 0 a < b c < d n. Find a closed formula for a n, as well as its ordinary generating
More informationMAT104: Fundamentals of Mathematics II Summary of Counting Techniques and Probability. Preliminary Concepts, Formulas, and Terminology
MAT104: Fundamentals of Mathematics II Summary of Counting Techniques and Probability Preliminary Concepts, Formulas, and Terminology Meanings of Basic Arithmetic Operations in Mathematics Addition: Generally
More informationWeek 3-4: Permutations and Combinations
Week 3-4: Permutations and Combinations February 20, 2017 1 Two Counting Principles Addition Principle. Let S 1, S 2,..., S m be disjoint subsets of a finite set S. If S = S 1 S 2 S m, then S = S 1 + S
More informationDiscrete Mathematics with Applications MATH236
Discrete Mathematics with Applications MATH236 Dr. Hung P. Tong-Viet School of Mathematics, Statistics and Computer Science University of KwaZulu-Natal Pietermaritzburg Campus Semester 1, 2013 Tong-Viet
More informationCHAPTER 5 BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS
CHAPTER 5 BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS BASIC CONCEPTS OF PERM UTATIONS AND COM BINATIONS LEARNING OBJECTIVES After reading this Chapter a student will be able to understand difference
More informationDiscrete Mathematics: Logic. Discrete Mathematics: Lecture 15: Counting
Discrete Mathematics: Logic Discrete Mathematics: Lecture 15: Counting counting combinatorics: the study of the number of ways to put things together into various combinations basic counting principles
More informationHOMEWORK ASSIGNMENT 5
HOMEWORK ASSIGNMENT 5 MATH 251, WILLIAMS COLLEGE, FALL 2006 Abstract. These are the instructor s solutions. 1. Big Brother The social security number of a person is a sequence of nine digits that are not
More informationElementary Combinatorics
184 DISCRETE MATHEMATICAL STRUCTURES 7 Elementary Combinatorics 7.1 INTRODUCTION Combinatorics deals with counting and enumeration of specified objects, patterns or designs. Techniques of counting are
More information2. Nine points are distributed around a circle in such a way that when all ( )
1. How many circles in the plane contain at least three of the points (0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)? Solution: There are ( ) 9 3 = 8 three element subsets, all
More informationNON-OVERLAPPING PERMUTATION PATTERNS. To Doron Zeilberger, for his Sixtieth Birthday
NON-OVERLAPPING PERMUTATION PATTERNS MIKLÓS BÓNA Abstract. We show a way to compute, to a high level of precision, the probability that a randomly selected permutation of length n is nonoverlapping. As
More informationMathematics Competition Practice Session 6. Hagerstown Community College: STEM Club November 20, :00 pm - 1:00 pm STC-170
2015-2016 Mathematics Competition Practice Session 6 Hagerstown Community College: STEM Club November 20, 2015 12:00 pm - 1:00 pm STC-170 1 Warm-Up (2006 AMC 10B No. 17): Bob and Alice each have a bag
More informationChapter 7. Intro to Counting
Chapter 7. Intro to Counting 7.7 Counting by complement 7.8 Permutations with repetitions 7.9 Counting multisets 7.10 Assignment problems: Balls in bins 7.11 Inclusion-exclusion principle 7.12 Counting
More informationRosen, Discrete Mathematics and Its Applications, 6th edition Extra Examples
Rosen, Discrete Mathematics and Its Applications, 6th edition Extra Examples Section 6.1 An Introduction to Discrete Probability Page references correspond to locations of Extra Examples icons in the textbook.
More informationQuestion No: 1 If you join all the vertices of a heptagon, how many quadrilaterals will you get?
Volume: 427 Questions Question No: 1 If you join all the vertices of a heptagon, how many quadrilaterals will you get? A. 72 B. 36 C. 25 D. 35 E. 120 Question No: 2 Four students have to be chosen 2 girls
More informationFoundations of Computing Discrete Mathematics Solutions to exercises for week 12
Foundations of Computing Discrete Mathematics Solutions to exercises for week 12 Agata Murawska (agmu@itu.dk) November 13, 2013 Exercise (6.1.2). A multiple-choice test contains 10 questions. There are
More informationMAT 243 Final Exam SOLUTIONS, FORM A
MAT 243 Final Exam SOLUTIONS, FORM A 1. [10 points] Michael Cow, a recent graduate of Arizona State, wants to put a path in his front yard. He sets this up as a tiling problem of a 2 n rectangle, where
More informationMATH 351 Fall 2009 Homework 1 Due: Wednesday, September 30
MATH 51 Fall 2009 Homework 1 Due: Wednesday, September 0 Problem 1. How many different letter arrangements can be made from the letters BOOKKEEPER. This is analogous to one of the problems presented in
More informationIntroduction. Firstly however we must look at the Fundamental Principle of Counting (sometimes referred to as the multiplication rule) which states:
Worksheet 4.11 Counting Section 1 Introduction When looking at situations involving counting it is often not practical to count things individually. Instead techniques have been developed to help us count
More informationJong C. Park Computer Science Division, KAIST
Jong C. Park Computer Science Division, KAIST Today s Topics Basic Principles Permutations and Combinations Algorithms for Generating Permutations Generalized Permutations and Combinations Binomial Coefficients
More informationMath Fall 2011 Exam 2 Solutions - November 1, 2011
Math 365 - Fall 011 Exam Solutions - November 1, 011 NAME: STUDENT ID: This is a closed-book and closed-note examination. Calculators are not allowed. Please show all your work. Use only the paper provided.
More informationContents 2.1 Basic Concepts of Probability Methods of Assigning Probabilities Principle of Counting - Permutation and Combination 39
CHAPTER 2 PROBABILITY Contents 2.1 Basic Concepts of Probability 38 2.2 Probability of an Event 39 2.3 Methods of Assigning Probabilities 39 2.4 Principle of Counting - Permutation and Combination 39 2.5
More informationPERMUTATIONS AND COMBINATIONS
8 PERMUTATIONS AND COMBINATIONS FUNDAMENTAL PRINCIPLE OF COUNTING Multiplication Principle : If an operation can be performed in 'm' different ways; following which a second operation can be performed
More informationTopics to be covered
Basic Counting 1 Topics to be covered Sum rule, product rule, generalized product rule Permutations, combinations Binomial coefficients, combinatorial proof Inclusion-exclusion principle Pigeon Hole Principle
More informationMathematical Foundations of Computer Science Lecture Outline August 30, 2018
Mathematical Foundations of omputer Science Lecture Outline ugust 30, 2018 ounting ounting is a part of combinatorics, an area of mathematics which is concerned with the arrangement of objects of a set
More informationPermutation and Combination
BANKERSWAY.COM Permutation and Combination Permutation implies arrangement where order of things is important. It includes various patterns like word formation, number formation, circular permutation etc.
More informationSolutions for Exam I, Math 10120, Fall 2016
Solutions for Exam I, Math 10120, Fall 2016 1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {1, 2, 3} B = {2, 4, 6, 8, 10}. C = {4, 5, 6, 7, 8}. Which of the following sets is equal to (A B) C? {1, 2, 3,
More informationReview I. October 14, 2008
Review I October 14, 008 If you put n + 1 pigeons in n pigeonholes then at least one hole would have more than one pigeon. If n(r 1 + 1 objects are put into n boxes, then at least one of the boxes contains
More informationCOMBINATORIAL PROBABILITY
COMBINATORIAL PROBABILITY Question 1 (**+) The Oakwood Jogging Club consists of 7 men and 6 women who go for a 5 mile run every Thursday. It is decided that a team of 8 runners would be picked at random
More informationCombinatorics: The Fine Art of Counting
Combinatorics: The Fine Art of Counting Lecture Notes Counting 101 Note to improve the readability of these lecture notes, we will assume that multiplication takes precedence over division, i.e. A / B*C
More informationProbability and Counting Techniques
Probability and Counting Techniques Diana Pell (Multiplication Principle) Suppose that a task consists of t choices performed consecutively. Suppose that choice 1 can be performed in m 1 ways; for each
More informationMath236 Discrete Maths with Applications
Math236 Discrete Maths with Applications P. Ittmann UKZN, Pietermaritzburg Semester 1, 2012 Ittmann (UKZN PMB) Math236 2012 1 / 43 The Multiplication Principle Theorem Let S be a set of k-tuples (s 1,
More informationTwenty-sixth Annual UNC Math Contest First Round Fall, 2017
Twenty-sixth Annual UNC Math Contest First Round Fall, 07 Rules: 90 minutes; no electronic devices. The positive integers are,,,,.... Find the largest integer n that satisfies both 6 < 5n and n < 99..
More informationCombinatorics and Intuitive Probability
Chapter Combinatorics and Intuitive Probability The simplest probabilistic scenario is perhaps one where the set of possible outcomes is finite and these outcomes are all equally likely. A subset of the
More informationCombinatorics (Part II)
Combinatorics (Part II) BEGINNERS 02/08/2015 Warm-Up (a) How many five-digit numbers are there? (b) How many are odd? (c) How many are odd and larger than 30,000? (d) How many have only odd digits? (e)
More informationIntroduction to Mathematical Reasoning, Saylor 111
Here s a game I like plying with students I ll write a positive integer on the board that comes from a set S You can propose other numbers, and I tell you if your proposed number comes from the set Eventually
More information# 1. As shown, the figure has been divided into three identical parts: red, blue, and green. The figures are identical because the blue and red
# 1. As shown, the figure has been divided into three identical parts: red, blue, and green. The figures are identical because the blue and red figures are already in the correct orientation, and the green
More informationNon-overlapping permutation patterns
PU. M. A. Vol. 22 (2011), No.2, pp. 99 105 Non-overlapping permutation patterns Miklós Bóna Department of Mathematics University of Florida 358 Little Hall, PO Box 118105 Gainesville, FL 326118105 (USA)
More informationChapter 1. Probability
Chapter 1. Probability 1.1 Basic Concepts Scientific method a. For a given problem, we define measures that explains the problem well. b. Data is collected with observation and the measures are calculated.
More informationCOUNTING TECHNIQUES. Prepared by Engr. JP Timola Reference: Discrete Math by Kenneth H. Rosen
COUNTING TECHNIQUES Prepared by Engr. JP Timola Reference: Discrete Math by Kenneth H. Rosen COMBINATORICS the study of arrangements of objects, is an important part of discrete mathematics. Counting Introduction
More informationFinite Math - Fall 2016
Finite Math - Fall 206 Lecture Notes - /28/206 Section 7.4 - Permutations and Combinations There are often situations in which we have to multiply many consecutive numbers together, for example, in examples
More informationBlock 1 - Sets and Basic Combinatorics. Main Topics in Block 1:
Block 1 - Sets and Basic Combinatorics Main Topics in Block 1: A short revision of some set theory Sets and subsets. Venn diagrams to represent sets. Describing sets using rules of inclusion. Set operations.
More informationcode V(n,k) := words module
Basic Theory Distance Suppose that you knew that an English word was transmitted and you had received the word SHIP. If you suspected that some errors had occurred in transmission, it would be impossible
More informationCounting Things Solutions
Counting Things Solutions Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles March 7, 006 Abstract These are solutions to the Miscellaneous Problems in the Counting Things article at:
More informationWeek 6: Advance applications of the PIE. 17 and 19 of October, 2018
(1/22) MA284 : Discrete Mathematics Week 6: Advance applications of the PIE http://www.maths.nuigalway.ie/ niall/ma284 17 and 19 of October, 2018 1 Stars and bars 2 Non-negative integer inequalities 3
More informationCompound Probability. Set Theory. Basic Definitions
Compound Probability Set Theory A probability measure P is a function that maps subsets of the state space Ω to numbers in the interval [0, 1]. In order to study these functions, we need to know some basic
More informationIB HL Mathematics Homework 2014
IB HL Mathematics Homework Counting, Binomial Theorem Solutions 1) How many permutations are there of the letters MATHEMATICS? Using the permutation formula, we get 11!/(2!2!2!), since there are 2 M's,
More informationReading 14 : Counting
CS/Math 240: Introduction to Discrete Mathematics Fall 2015 Instructors: Beck Hasti, Gautam Prakriya Reading 14 : Counting In this reading we discuss counting. Often, we are interested in the cardinality
More informationOlympiad Combinatorics. Pranav A. Sriram
Olympiad Combinatorics Pranav A. Sriram August 2014 Chapter 2: Algorithms - Part II 1 Copyright notices All USAMO and USA Team Selection Test problems in this chapter are copyrighted by the Mathematical
More informationSolutions to Exercises on Page 86
Solutions to Exercises on Page 86 #. A number is a multiple of, 4, 5 and 6 if and only if it is a multiple of the greatest common multiple of, 4, 5 and 6. The greatest common multiple of, 4, 5 and 6 is
More informationThe Pigeonhole Principle
The Pigeonhole Principle Some Questions Does there have to be two trees on Earth with the same number of leaves? How large of a set of distinct integers between 1 and 200 is needed to assure that two numbers
More information6.1.1 The multiplication rule
6.1.1 The multiplication rule 1. There are 3 routes joining village A and village B and 4 routes joining village B and village C. Find the number of different ways of traveling from village A to village
More informationCounting Subsets with Repetitions. ICS 6C Sandy Irani
Counting Subsets with Repetitions ICS 6C Sandy Irani Multi-sets A Multi-set can have more than one copy of an item. Order doesn t matter The number of elements of each type does matter: {1, 2, 2, 2, 3,
More information(1). We have n different elements, and we would like to arrange r of these elements with no repetition, where 1 r n.
BASIC KNOWLEDGE 1. Two Important Terms (1.1). Permutations A permutation is an arrangement or a listing of objects in which the order is important. For example, if we have three numbers 1, 5, 9, there
More informationCounting in Algorithms
Counting Counting in Algorithms How many comparisons are needed to sort n numbers? How many steps to compute the GCD of two numbers? How many steps to factor an integer? Counting in Games How many different
More informationChapter 2. Permutations and Combinations
2. Permutations and Combinations Chapter 2. Permutations and Combinations In this chapter, we define sets and count the objects in them. Example Let S be the set of students in this classroom today. Find
More informationMGF 1106: Exam 2 Solutions
MGF 1106: Exam 2 Solutions 1. (15 points) A coin and a die are tossed together onto a table. a. What is the sample space for this experiment? For example, one possible outcome is heads on the coin and
More informationMAT 115: Finite Math for Computer Science Problem Set 5
MAT 115: Finite Math for Computer Science Problem Set 5 Out: 04/10/2017 Due: 04/17/2017 Instructions: I leave plenty of space on each page for your computation. If you need more sheet, please attach your
More information11 + Entrance Examination Sample Paper 2 Mathematics Total Marks: 100 Time allowed:1 hour
11 + Entrance Examination Sample Paper 2 Mathematics Total Marks: 100 Time allowed:1 hour Information for parents: This sample paper has been created for children who are embarking on the 11+ exam. The
More informationSolutions to Exercises Chapter 6: Latin squares and SDRs
Solutions to Exercises Chapter 6: Latin squares and SDRs 1 Show that the number of n n Latin squares is 1, 2, 12, 576 for n = 1, 2, 3, 4 respectively. (b) Prove that, up to permutations of the rows, columns,
More informationRANDOM EXPERIMENTS AND EVENTS
Random Experiments and Events 18 RANDOM EXPERIMENTS AND EVENTS In day-to-day life we see that before commencement of a cricket match two captains go for a toss. Tossing of a coin is an activity and getting
More information50 Counting Questions
50 Counting Questions Prob-Stats (Math 3350) Fall 2012 Formulas and Notation Permutations: P (n, k) = n!, the number of ordered ways to permute n objects into (n k)! k bins. Combinations: ( ) n k = n!,
More informationIvan Guo. Broken bridges There are thirteen bridges connecting the banks of River Pluvia and its six piers, as shown in the diagram below:
Ivan Guo Welcome to the Australian Mathematical Society Gazette s Puzzle Corner No. 20. Each issue will include a handful of fun, yet intriguing, puzzles for adventurous readers to try. The puzzles cover
More informationSimple Counting Problems
Appendix F Counting Principles F1 Appendix F Counting Principles What You Should Learn 1 Count the number of ways an event can occur. 2 Determine the number of ways two or three events can occur using
More information(i) Understanding of the characteristics of linear-phase finite impulse response (FIR) filters
FIR Filter Design Chapter Intended Learning Outcomes: (i) Understanding of the characteristics of linear-phase finite impulse response (FIR) filters (ii) Ability to design linear-phase FIR filters according
More informationCombinatorics. PIE and Binomial Coefficients. Misha Lavrov. ARML Practice 10/20/2013
Combinatorics PIE and Binomial Coefficients Misha Lavrov ARML Practice 10/20/2013 Warm-up Po-Shen Loh, 2013. If the letters of the word DOCUMENT are randomly rearranged, what is the probability that all
More informationCOUNTING AND PROBABILITY
CHAPTER 9 COUNTING AND PROBABILITY It s as easy as 1 2 3. That s the saying. And in certain ways, counting is easy. But other aspects of counting aren t so simple. Have you ever agreed to meet a friend
More informationProblem Set 8 Solutions R Y G R R G
6.04/18.06J Mathematics for Computer Science April 5, 005 Srini Devadas and Eric Lehman Problem Set 8 Solutions Due: Monday, April 11 at 9 PM in Room 3-044 Problem 1. An electronic toy displays a 4 4 grid
More informationPermutations and Combinations. Quantitative Aptitude & Business Statistics
Permutations and Combinations Statistics The Fundamental Principle of If there are Multiplication n 1 ways of doing one operation, n 2 ways of doing a second operation, n 3 ways of doing a third operation,
More informationMAT104: Fundamentals of Mathematics II Counting Techniques Class Exercises Solutions
MAT104: Fundamentals of Mathematics II Counting Techniques Class Exercises Solutions 1. Appetizers: Salads: Entrées: Desserts: 2. Letters: (A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U,
More informationChapter 1. Probability
Chapter 1. Probability 1.1 Basic Concepts Scientific method a. For a given problem, we define measures that explains the problem well. b. Data is collected with observation and the measures are calculated.
More informationCISC 1400 Discrete Structures
CISC 1400 Discrete Structures Chapter 6 Counting CISC1400 Yanjun Li 1 1 New York Lottery New York Mega-million Jackpot Pick 5 numbers from 1 56, plus a mega ball number from 1 46, you could win biggest
More informationMathematics. Programming
Mathematics for the Digital Age and Programming in Python >>> Second Edition: with Python 3 Maria Litvin Phillips Academy, Andover, Massachusetts Gary Litvin Skylight Software, Inc. Skylight Publishing
More informationMathematics. (www.tiwariacademy.com) (Chapter 7) (Permutations and Combinations) (Class XI) Exercise 7.3
Question 1: Mathematics () Exercise 7.3 How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated? Answer 1: 3-digit numbers have to be formed using the digits 1 to 9. Here,
More informationPermutations and codes:
Hamming distance Permutations and codes: Polynomials, bases, and covering radius Peter J. Cameron Queen Mary, University of London p.j.cameron@qmw.ac.uk International Conference on Graph Theory Bled, 22
More informationLecture 18 - Counting
Lecture 18 - Counting 6.0 - April, 003 One of the most common mathematical problems in computer science is counting the number of elements in a set. This is often the core difficulty in determining a program
More informationThere are three types of mathematicians. Those who can count and those who can t.
1 Counting There are three types of mathematicians. Those who can count and those who can t. 1.1 Orderings The details of the question always matter. So always take a second look at what is being asked
More information12. 6 jokes are minimal.
Pigeonhole Principle Pigeonhole Principle: When you organize n things into k categories, one of the categories has at least n/k things in it. Proof: If each category had fewer than n/k things in it then
More informationDiscrete Structures for Computer Science
Discrete Structures for Computer Science William Garrison bill@cs.pitt.edu 6311 Sennott Square Lecture #22: Generalized Permutations and Combinations Based on materials developed by Dr. Adam Lee Counting
More information(i) Understanding of the characteristics of linear-phase finite impulse response (FIR) filters
FIR Filter Design Chapter Intended Learning Outcomes: (i) Understanding of the characteristics of linear-phase finite impulse response (FIR) filters (ii) Ability to design linear-phase FIR filters according
More informationPRMO Official Test / Solutions
Date: 19 Aug 2018 PRMO Official Test - 2018 / Solutions 1. 17 ANSWERKEY 1. 17 2. 8 3. 70 4. 12 5. 84 6. 18 7. 14 8. 80 9. 81 10. 24 11. 29 12. 88 13. 24 14. 19 15. 21 16. 55 17. 30 18. 16 19. 33 20. 17
More informationCombinatorics: The Fine Art of Counting
Combinatorics: The Fine Art of Counting Week 6 Lecture Notes Discrete Probability Note Binomial coefficients are written horizontally. The symbol ~ is used to mean approximately equal. Introduction and
More informationChapter 3 PRINCIPLE OF INCLUSION AND EXCLUSION
Chapter 3 PRINCIPLE OF INCLUSION AND EXCLUSION 3.1 The basics Consider a set of N obects and r properties that each obect may or may not have each one of them. Let the properties be a 1,a,..., a r. Let
More informationPHYSICS 140A : STATISTICAL PHYSICS HW ASSIGNMENT #1 SOLUTIONS
PHYSICS 40A : STATISTICAL PHYSICS HW ASSIGNMENT # SOLUTIONS () The information entropy of a distribution {p n } is defined as S n p n log 2 p n, where n ranges over all possible configurations of a given
More information5. (1-25 M) How many ways can 4 women and 4 men be seated around a circular table so that no two women are seated next to each other.
A.Miller M475 Fall 2010 Homewor problems are due in class one wee from the day assigned (which is in parentheses. Please do not hand in the problems early. 1. (1-20 W A boo shelf holds 5 different English
More informationMath 3012 Applied Combinatorics Lecture 2
August 20, 2015 Math 3012 Applied Combinatorics Lecture 2 William T. Trotter trotter@math.gatech.edu The Road Ahead Alert The next two to three lectures will be an integrated approach to material from
More informationMath 475, Problem Set #3: Solutions
Math 475, Problem Set #3: Solutions A. Section 3.6, problem 1. Also: How many of the four-digit numbers being considered satisfy (a) but not (b)? How many satisfy (b) but not (a)? How many satisfy neither
More informationMATH 215 DISCRETE MATHEMATICS INSTRUCTOR: P. WENG
MATH DISCRETE MATHEMATICS INSTRUCTOR: P. WENG Counting and Probability Suggested Problems Basic Counting Skills, Inclusion-Exclusion, and Complement. (a An office building contains 7 floors and has 7 offices
More informationSTAT 430/510 Probability Lecture 3: Space and Event; Sample Spaces with Equally Likely Outcomes
STAT 430/510 Probability Lecture 3: Space and Event; Sample Spaces with Equally Likely Outcomes Pengyuan (Penelope) Wang May 25, 2011 Review We have discussed counting techniques in Chapter 1. (Principle
More informationUNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST
UNIVERSITY OF NORTHERN COLORADO MATHEMATICS CONTEST First Round For all Colorado Students Grades 7-12 October 31, 2009 You have 90 minutes no calculators allowed The average of n numbers is their sum divided
More informationChapter 2 Math
Chapter 2 Math 3201 1 Chapter 2: Counting Methods: Solving problems that involve the Fundamental Counting Principle Understanding and simplifying expressions involving factorial notation Solving problems
More informationPermutations and Combinations
Permutations and Combinations Introduction Permutations and combinations refer to number of ways of selecting a number of distinct objects from a set of distinct objects. Permutations are ordered selections;
More informationWith Question/Answer Animations. Chapter 6
With Question/Answer Animations Chapter 6 Chapter Summary The Basics of Counting The Pigeonhole Principle Permutations and Combinations Binomial Coefficients and Identities Generalized Permutations and
More informationSolutions to Problem Set 7
Massachusetts Institute of Technology 6.4J/8.6J, Fall 5: Mathematics for Computer Science November 9 Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld revised November 3, 5, 3 minutes Solutions to Problem
More information