MAT 243 Final Exam SOLUTIONS, FORM A
|
|
- Charlene Booker
- 5 years ago
- Views:
Transcription
1 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 the tiles are 2 1 rectangles, which come in two colors (maroon and gold). Let T n be the number of tilings such that there are no vertically-oriented maroon tiles. Find (but do not solve) a recurrence for T n. Solution: During Fall 2014, I put a problem similar to this on Test 1 (where you had to find a recurrence relation), and I will not put one on the Final Exam. If you want to work on it, though, you should end up with the recurrence T n = T n T n 2, n 3 T 1 = 1 T 2 = 5 For the rest of the problems, I WILL explain the solutions, though! 2. [10 points] Solve the linear homogeneous constant coefficient recurrence relation below. a n 6a n 1 + 9a n 2 = 0 a 0 = 2 a 1 = 12 Solution: Letting a n = r n, we obtain the following characteristic (auxiliary) equation: r 2 6r + 9 = 0 (r 3)(r 3) = 0 which has 3 as a solution with multiplicity 2. Hence, the general form for a n is a n = C 1 3 n + C 2 n 3 n. Substituting n = 0 and n = 1 gives conditions that are necessary for C 1 and C 2 : n = 0 : 2 = C C = C 1 n = 1 : 12 = C C = C 2 so C 1 = 2 and C 2 = 2. Thus, a n = 2 3 n + 2 n 3 n. 2
2 MAT 243 Final Exam SOLUTIONS, FORM A 3. Recall that a standard deck of playing card consists of four suits (,,, ) and thirteen ranks (or kinds ) (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K), and that every card has one rank and one suit. The game of Rekop is played with a standard deck of playing cards, where each player receives four cards. The order that the cards are received does not matter. a. [5 points] How many hands are possible? Solution: Here, you are asked to choose r = 4 cards from a set of n = 52. The order of the cards does not matter, and you are not allowed to choose the same card more than once. Hence the number of hands is C(52, 4) = 270,725. b. [5 points] How many hands consist of two pairs (two cards of one rank, and two cards of another rank, such as )? Solution: One of these two pairs hands is determined uniquely once you choose: (a) the ranks of the pairs, (b) the suits of the higher pair, and (c) the suits of the lower pair. (The example hand will be a result of the answers {3, 5}, {, }, and {, }, in that order. Note that you cannot choose the ranks one at a time; in that case, the example hand could be obtained with the answers 3, 5, {, }, and {, }, as well as the answers 5, 3, {, }, and {, }.) In choosing the ranks, you are choosing r = 2 ranks from n = 13 possibilities. The order does not matter (otherwise, you would be overcounting), and repetition is not allowed, so the number of ways to answer (a) is C(13, 2). The number of ways to answer (b) and (c) is C(4, 2), because the order of the suits does not matter, and repetition is not allowed. The total number of two pairs hands is thus C(13, 2) C(4, 2) C(4, 2) = 2808, by the multiplication principle. c. [5 points] How many hands consist of three of a kind (three cards of one rank and one of another rank, such as A )? Solution: The sequence of decisions to be made to obtain a specific three of a kind is: (a) the rank of the three-of-a-kind, (b) the rank of the fourth card, (c) the suits of the three-of-a-kind, and (d) the suit of the fourth card. (The sample hand is a result of the answers 8, A, {,, } and.) The number of ways to answer (a) is 13, the number of ways to answer (b) is 12, the number of ways to answer (c) is C(4, 3), and the number of ways to answer (d) is 4 (which is also C(4, 1)). By the multiplication principle, the total number of three of a kind hands is C(4, 3) 4 = [10 points] Find the coefficient of x 7 in (2x 5) 70. Solution: According to the Binomial Theorem, (2x 5) 70 = C(70, k) (2x) k ( 5) 70 k = [C(70, k) 2 k ( 5) 70 k ] x k. k=0 We want the term with x 7, which occurs when k = 7. The coefficient of x 7 is the value of the expression inside brackets when k = 7, which is C(70, 7) 2 7 ( 5) 63 (which is exactly 16,636,341,193,532,189,208,781,346,678,733,825,683,593,750,000,000,000,000, in case you re interested.) k=0 3
3 MAT 243 Final Exam SOLUTIONS, FORM A 5. Let R = {(1, 2), (1, 3), (2, 1), (2, 2), (3, 4), (4, 4)}. a. [10 points] Calculate R R. Solution: (x, y) R R iff there is a z such that (x, z) R and (z, y) R. For each ordered pair (x, z) R below, the possible ordered pairs (z, y) R are listed right after it, and the resulting (x, y) R R is listed to the far right: Combining the ordered pairs at the far right, we get (1, 2) : (2, 1), (2, 2), so (1, 1), (1, 2) R R (1, 3) : (3, 4) (1, 4) R R (2, 1) : (1, 2), (1, 3) (2, 2), (2, 3) R R (2, 2) : (2, 1), (2, 2) (2, 1), (2, 2) R (3, 4) : (4, 4) (3, 4) R R (4, 4) : (4, 4) (4, 4) R R R R = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 3), (3, 4), (4, 4)}. b. [10 points] Which ordered pairs must be added to R so that the resulting relation is symmetric? Solution: In order for a relation R to be symmetric, you must have (y, x) R whenever (x, y) R. Given the ordered pairs already in R, that means you must also have the ordered pairs (2, 1), (3, 1), (1, 2), (2, 2), (4, 3), and (4, 4). The ordered pairs that need to be added are the ones not already in R, namely (3, 1), (4, 3). The resulting relation is symmetric, so we are done. (See Form B, though!) c. [10 points] Which ordered pairs must be REMOVED from R so that the resulting relation is irreflexive? Solution: A relation being irreflexive means (a, a) R for any a; thus, any such ordered pairs need to be removed. In this case, these are the ordered pairs (2, 2) and (4, 4). The resulting relation is irreflexive, so the answer is (2, 2), (4, 4). 4
4 MAT 243 Final Exam SOLUTIONS, FORM A 6. [10 points] Let D 70 = {1, 2, 5, 7, 10, 14, 35, 70} be the set of positive divisors of 70, and suppose that S is a subset of D 70. Prove that if S has at least five elements, then there are two elements of S whose product is 70. Solution: Set up four boxes in the following way: One box can have only 1 and 70 put into it; Another box can have only 2 and 35 put into it; Another box can have only 5 and 14 put into it; The final box can have only 7 and 10 put into it. Now choose a set S with five elements, where each element is an element of D 70. Distribute the elements of S among the boxes. The Pigeonhole Principle implies that some box has at least = numbers in it. These two numbers are elements of S, and their product is 70, by how the boxes were set up. Thus, the assertion has been proven. 7. [15 points] Use inclusion-exclusion to count how many functions from S = {a, b, c, d} to T = {1, 2, 3} which are NOT onto. Do this by letting F i be the set of functions f from S to T such that f(x) i for all x S and finding F 1 F 2 F 3. You will also need the following information: The number of functions from A to B is B A (So, for example, the number of functions from S to T is 3 5 = 243.) Solution: This is one of those problems that you reason your way through, one step at a time. First of all, you are asked to find F 1 F 2 F 3. Inclusion-Exclusion implies that F 1 F 2 F 3 = F 1 + F 2 + F 3 F 1 F 2 F 1 F 3 F 2 F 3 + F 1 F 2 F 3. There are seven things to calculate here, so we look at them one at a time. First of all, the sets F i all contain functions. That means if you choose an element f T i, then f is an assignment of one of the numbers 1, 2, or 3 to each element of S. For instance, a single function can be represented by the four equations f(a) = 1; f(b) = 2; f(c) = 1; f(d) = 2. This function happens not to be in F 1, because f(a) = 1. It is not in F 2, either, because f(b) = 2. However, it IS in F 3, because none of f(a), f(b), f(c), f(d) equals 3. Now, F 1 is the set of all functions f from S to {1, 2, 3} such that f(x) 1 for any x S. This means f is really a function from S to {2, 3}. The fact presented above allows us to calculate F 1 ; it is just {2, 3} S = 2 4 = 16. Similarly, F 2 is the set of all functions from S to {1, 3}; it also has size 2 4 = 16. Lastly F 3 is the set of all functions from S to {1, 2}, and the number of such functions is 2 4 = 16 as well. So far, so good. Now for F 1 F 2 ; what does this mean? This is the set of all functions where f(x) 1 and f(x) 2. That just means you MUST assign 3 to every element of S! Thus, F 1 F 2 consists of just one function f, namely the one which has the following properties: f(a) = 3; f(b) = 3; f(c) = 3; f(d) = 3. Thus, F 1 F 2 = 1. (You can also use the formula that was supplied to obtain this number.) Now, F 1 F 3 = 1 as well, because (following the reasoning above), there is only one element f in F 1 F 3, the function that assigns 2 to every element of S. Similarly, F 2 F 3 = 1. Now for the final lap: What is F 1 F 2 F 3? Well, an element of this set is a function f which assigns one of 1, 2, or 3 to each element of S, except that f(x) 1, f(x) 2, and f(x) 3. Is this possible? No! So F 1 F 2 F 3 is empty, and F 1 F 2 F 3 = 0. Now you have enough data to finish the problem: F 1 F 2 F 3 = = 45. 5
5 MAT 243 Final Exam SOLUTIONS, FORM 1 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 the tiles are 2 1 rectangles, which come in two colors (maroon and gold). Let T n be the number of tilings such that there are no horizonally-oriented maroon tiles. Find (but do not solve) a recurrence for T n. Solution: During Fall 2014, I put a problem similar to this on Test 1 (where you had to find a recurrence relation), and I will not put one on the Final Exam. If you want to work on it, though, you should end up with the recurrence T n = 2 T n 1 + T n 2, n 3 T 1 = 2 T 2 = 5 For the rest of the problems, I WILL explain the solutions, though! 2. [10 points] Solve the linear homogeneous constant coefficient recurrence relation below. a n 8a n a n 2 = 0 a 0 = 1 a 1 = 8 Solution: Letting a n = r n, we obtain the following characteristic (auxiliary) equation: r 2 8r + 16 = 0 (r 4)(r 4) = 0 which has 4 as a solution with multiplicity 2. Hence, the general form for a n is a n = C 1 4 n + C 2 n 4 n. Substituting n = 0 and n = 1 gives conditions that are necessary for C 1 and C 2 : n = 0 : 1 = C C = C 1 n = 1 : 8 = C C = C 2 so C 1 = 1 and C 2 = 3. Thus, a n = 1 4 n + 3 n 4 n. 2
6 MAT 243 Final Exam SOLUTIONS, FORM 1 3. Recall that a standard deck of playing card consists of four suits (,,, ) and thirteen ranks (or kinds ) (A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K), and that every card has one rank and one suit. The game of Rekop is played with a standard deck of playing cards, where each player receives six cards. The order that the cards are received does not matter. a. [5 points] How many hands are possible? Solution: Here, you are asked to choose r = 6 cards from a set of n = 52. The order of the cards does not matter, and you are not allowed to choose the same card more than once. Hence the number of hands is C(52, 6) = 20,358,520. b. [5 points] How many hands consist of three pairs (two cards of one rank, two cards of another rank, and two cards of yet another rank, such as )? Solution: One of these three pairs hands is determined uniquely once you choose: (a) the ranks of the pairs, (b) the suits of the higher pair, (c) the suits of the middle pair, and (d) the suits of the lower pair. (The example hand will be a result of the answers {3, 5, 10}, {, }, {, }, and {, }, in that order. Note that you cannot choose the ranks one at a time; in that case, the example hand could be obtained with the answers 3, 5, 10, {, }, {, }, and {, }, as well as the answers 5, 3, 10, {, }, {, }, and {, }.) In choosing the ranks, you are choosing r = 3 ranks from n = 13 possibilities. The order does not matter (otherwise, you would be overcounting), and repetition is not allowed, so the number of ways to answer (a) is C(13, 3). The number of ways to answer (b), (c), and (d) is C(4, 2), because the order of the suits does not matter, and repetition is not allowed. The total number of two pairs hands is thus C(13, 3) C(4, 2) C(4, 2) C(4, 2) = 61,776, by the multiplication principle. c. [5 points] How many hands consist of four of a kind (four cards of one rank and two of other ranks, such as A K )? Solution: The sequence of decisions to be made to obtain a specific four of a kind is: (a) the rank of the four-of-a-kind, (b) the ranks of the remaining cards, (c) the suits of the four-of-a-kind, (d) the suit of the higher card which is not a four-of-a-kind, and (e) the suit of the lower card which is not a four-of-a-kind. (The sample hand is a result of the answers 8, {A, K}, {,,, },, and.) The number of ways to answer (a) is 13, the number of ways to answer (b) is C(12, 2), the number of ways to answer (c) is C(4, 4) (which is just 1), and the number of ways to answer (d) and (e) is 4 (which is also C(4, 1)). By the multiplication principle, the total number of four of a kind hands is 13 C(12, 3) = 13, [10 points] Find the coefficient of x 6 in (3x 4) 60. Solution: According to the Binomial Theorem, (3x 4) 60 = C(60, k) (3x) k ( 4) 60 k = [C(60, k) 3 k ( 4) 60 k ] x k. k=0 We want the term with x 6, which occurs when k = 6. The coefficient of x 6 is the value of the expression inside brackets when k = 6, which is C(60, 6) 3 6 ( 4) 54 (which is exactly in case you re interested.) k=0 11,843,808,898,125,555,369,579,096,388,397,076,987,248,640, 3
7 MAT 243 Final Exam SOLUTIONS, FORM 1 5. Let R = {(1, 2), (2, 2), (2, 3), (2, 4), (3, 2)}. a. [10 points] Calculate R R. Solution: (x, y) R R iff there is a z such that (x, z) R and (z, y) R. For each ordered pair (x, z) R below, the possible ordered pairs (z, y) R are listed right after it, and the resulting (x, y) R R is listed to the far right: (1, 2) : (2, 2), (2, 3), (2, 4), so (1, 2), (1, 3), (1, 4) R R (2, 2) : (2, 2), (2, 3), (2, 4) (2, 2), (2, 3), (2, 4) R R (2, 3) : (3, 2) (2, 2) R R (2, 4) : (none) Combining the ordered pairs at the far right, we get (3, 2) : (2, 2), (2, 3), (2, 4) (3, 2), (3, 3), (3, 4) R R R R = {(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4)}. b. [10 points] Which ordered pairs must be added to R so that the resulting relation is transitive? Solution: In order for a relation R to be transitive, you must have (x, z) R whenever (x, y) R and (y, z) R. In other words, the elements of R R not already in R need to be added to R. This means (1, 3), (1, 4), (3, 3), and (3, 4) need to be added to R. BUT... and this is a big but... adding these elements will NOT NECESSARILY make the new relation transitive! In this case, if you look at all possible combinations, you see that adding these four ordered pairs will make a transitive relation, so the answer is (1, 3), (1, 4), (3, 3), (3, 4), but there are cases where the new relation will not be transitive. For example, if we had R = {(a, b), (b, c), (c, d)}, the first paragraph would tell us to add the ordered pairs (a, c) and (b, d) to get S = {(a, b), (a, c), (b, c), (b, d), (c, d)}. However, S is not transititive: (a, b) S and (b, d) S, but (a, d) S. So (a, d) would also have to be added to S to get T, and T turns out to be transitive. c. [10 points] Which ordered pairs must be REMOVED from R so that the resulting relation is antisymmetric? Solution: A relation being irreflexive means: You can only have (a, b) R and (b, a) R only if a = b; if you can find a b such that (a, b), (b, a) R, then you must remove one (or both) of them to get an antisymmetric relation. In this case, the relation is already antisymmetric, so the answer is none. (If antisymmetric was changed to asymmetric, you would have to remove any ordered pairs (a, a) which are in R, though. Hence, you would have to remove (2, 2) and only that ordered pair to get an asymmetric relation.) 4
8 MAT 243 Final Exam SOLUTIONS, FORM 1 6. [10 points] Let D 42 = {1, 2, 3, 6, 7, 14, 21, 42} be the set of positive divisors of 70, and suppose that S is a subset of D 42. Prove that if S has at least five elements, then there are two elements of S whose product is 42. Solution: Set up four boxes in the following way: One box can have only 1 and 42 put into it; Another box can have only 2 and 21 put into it; Another box can have only 3 and 14 put into it; The final box can have only 6 and 7 put into it. Now choose a set S with five elements, where each element is an element of D 42. Distribute the elements of S among the boxes. The Pigeonhole Principle implies that some box has at least = numbers in it. These two numbers are elements of S, and their product is 42, by how the boxes were set up. Thus, the assertion has been proven. 7. [15 points] Use inclusion-exclusion to count how many functions from S = {a, b, c, d, e} to T = {1, 2, 3} which are NOT onto. Do this by letting F i be the set of functions f from S to T such that f(x) i for all x S and finding F 1 F 2 F 3. You will also need the following information: The number of functions from A to B is B A (So, for example, the number of functions from S to T is 3 5 = 243.) Solution: This is one of those problems that you reason your way through, one step at a time. First of all, you are asked to find F 1 F 2 F 3. Inclusion-Exclusion implies that F 1 F 2 F 3 = F 1 + F 2 + F 3 F 1 F 2 F 1 F 3 F 2 F 3 + F 1 F 2 F 3. There are seven things to calculate here, so we look at them one at a time. First of all, the sets F i all contain functions. That means if you choose an element f T i, then f is an assignment of one of the numbers 1, 2, or 3 to each element of S. For instance, a single function can be represented by the five equations f(a) = 1; f(b) = 2; f(c) = 1; f(d) = 2; f(e) = 2. This function happens not to be in F 1, because f(a) = 1. It is not in F 2, either, because f(b) = 2. However, it IS in F 3, because none of f(a), f(b), f(c), f(d), f(e) equals 3. Now, F 1 is the set of all functions f from S to {1, 2, 3} such that f(x) 1 for any x S. This means f is really a function from S to {2, 3}. The fact presented above allows us to calculate F 1 ; it is just {2, 3} S = 2 5 = 32. Similarly, F 2 is the set of all functions from S to {1, 3}; it also has size 2 5 = 32. Lastly F 3 is the set of all functions from S to {1, 2}, and the number of such functions is 2 5 = 32 as well. So far, so good. Now for F 1 F 2 ; what does this mean? This is the set of all functions where f(x) 1 and f(x) 2. That just means you MUST assign 3 to every element of S! Thus, F 1 F 2 consists of just one function f, namely the one which has the following properties: f(a) = 3; f(b) = 3; f(c) = 3; f(d) = 3; f(e) = 3. Thus, F 1 F 2 = 1. (You can also use the formula that was supplied to obtain this number.) Now, F 1 F 3 = 1 as well, because (following the reasoning above), there is only one element f in F 1 F 3, the function that assigns 2 to every element of S. Similarly, F 2 F 3 = 1. Now for the final lap: What is F 1 F 2 F 3? Well, an element of this set is a function f which assigns one of 1, 2, or 3 to each element of S, except that f(x) 1, f(x) 2, and f(x) 3. Is this possible? No! So F 1 F 2 F 3 is empty, and F 1 F 2 F 3 = 0. Now you have enough data to finish the problem: F 1 F 2 F 3 = = 93. 5
CS100: 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 informationPrinciple of Inclusion-Exclusion Notes
Principle of Inclusion-Exclusion Notes The Principle of Inclusion-Exclusion (often abbreviated PIE is the following general formula used for finding the cardinality of a union of finite sets. Theorem 0.1.
More informationLECTURE 7: POLYNOMIAL CONGRUENCES TO PRIME POWER MODULI
LECTURE 7: POLYNOMIAL CONGRUENCES TO PRIME POWER MODULI 1. Hensel Lemma for nonsingular solutions Although there is no analogue of Lagrange s Theorem for prime power moduli, there is an algorithm for determining
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 informationDistribution of Aces Among Dealt Hands
Distribution of Aces Among Dealt Hands Brian Alspach 3 March 05 Abstract We provide details of the computations for the distribution of aces among nine and ten hold em hands. There are 4 aces and non-aces
More informationCMath 55 PROFESSOR KENNETH A. RIBET. Final Examination May 11, :30AM 2:30PM, 100 Lewis Hall
CMath 55 PROFESSOR KENNETH A. RIBET Final Examination May 11, 015 11:30AM :30PM, 100 Lewis Hall Please put away all books, calculators, cell phones and other devices. You may consult a single two-sided
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 informationIt is important that you show your work. The total value of this test is 220 points.
June 27, 2001 Your name It is important that you show your work. The total value of this test is 220 points. 1. (10 points) Use the Euclidean algorithm to solve the decanting problem for decanters of sizes
More informationn! = n(n 1)(n 2) 3 2 1
A Counting A.1 First principles If the sample space Ω is finite and the outomes are equally likely, then the probability measure is given by P(E) = E / Ω where E denotes the number of outcomes in the event
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 informationMAT3707. Tutorial letter 202/1/2017 DISCRETE MATHEMATICS: COMBINATORICS. Semester 1. Department of Mathematical Sciences MAT3707/202/1/2017
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
More information9.5 Counting Subsets of a Set: Combinations. Answers for Test Yourself
9.5 Counting Subsets of a Set: Combinations 565 H 35. H 36. whose elements when added up give the same sum. (Thanks to Jonathan Goldstine for this problem. 34. Let S be a set of ten integers chosen from
More informationFinal Exam, Math 6105
Final Exam, Math 6105 SWIM, June 29, 2006 Your name Throughout this test you must show your work. 1. Base 5 arithmetic (a) Construct the addition and multiplication table for the base five digits. (b)
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 informationSolution: This is sampling without repetition and order matters. Therefore
June 27, 2001 Your name It is important that you show your work. The total value of this test is 220 points. 1. (10 points) Use the Euclidean algorithm to solve the decanting problem for decanters of sizes
More informationCSE 312: Foundations of Computing II Quiz Section #2: Combinations, Counting Tricks (solutions)
CSE 312: Foundations of Computing II Quiz Section #2: Combinations, Counting Tricks (solutions Review: Main Theorems and Concepts Combinations (number of ways to choose k objects out of n distinct objects,
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 information1 = 3 2 = 3 ( ) = = = 33( ) 98 = = =
Math 115 Discrete Math Final Exam December 13, 2000 Your name It is important that you show your work. 1. Use the Euclidean algorithm to solve the decanting problem for decanters of sizes 199 and 98. In
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 informationThe Product Rule The Product Rule: A procedure can be broken down into a sequence of two tasks. There are n ways to do the first task and n
Chapter 5 Chapter Summary 5.1 The Basics of Counting 5.2 The Pigeonhole Principle 5.3 Permutations and Combinations 5.5 Generalized Permutations and Combinations Section 5.1 The Product Rule The Product
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 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 informationand problem sheet 7
1-18 and 15-151 problem sheet 7 Solutions to the following five exercises and optional bonus problem are to be submitted through gradescope by 11:30PM on Friday nd November 018. Problem 1 Let A N + and
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 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 informationCSE 312: Foundations of Computing II Quiz Section #2: Inclusion-Exclusion, Pigeonhole, Introduction to Probability (solutions)
CSE 31: Foundations of Computing II Quiz Section #: Inclusion-Exclusion, Pigeonhole, Introduction to Probability (solutions) Review: Main Theorems and Concepts Binomial Theorem: x, y R, n N: (x + y) n
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 informationCounting. Chapter 6. With Question/Answer Animations
. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education. Counting Chapter
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 informationMaxima and Minima. Terminology note: Do not confuse the maximum f(a, b) (a number) with the point (a, b) where the maximum occurs.
10-11-2010 HW: 14.7: 1,5,7,13,29,33,39,51,55 Maxima and Minima In this very important chapter, we describe how to use the tools of calculus to locate the maxima and minima of a function of two variables.
More informationarxiv: v1 [math.co] 24 Nov 2018
The Problem of Pawns arxiv:1811.09606v1 [math.co] 24 Nov 2018 Tricia Muldoon Brown Georgia Southern University Abstract Using a bijective proof, we show the number of ways to arrange a maximum number of
More informationSolutions to Problem Set 6 - Fall 2008 Due Tuesday, Oct. 21 at 1:00
18.781 Solutions to Problem Set 6 - Fall 008 Due Tuesday, Oct. 1 at 1:00 1. (Niven.8.7) If p 3 is prime, how many solutions are there to x p 1 1 (mod p)? How many solutions are there to x p 1 (mod p)?
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 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 informationUCSD CSE 21, Spring 2014 [Section B00] Mathematics for Algorithm and System Analysis
UCSD CSE 21, Spring 2014 [Section B00] Mathematics for Algorithm and System Analysis Lecture 3 Class URL: http://vlsicad.ucsd.edu/courses/cse21-s14/ Lecture 3 Notes Goal for today: CL Section 3 Subsets,
More informationBMT 2018 Combinatorics Test Solutions March 18, 2018
. Bob has 3 different fountain pens and different ink colors. How many ways can he fill his fountain pens with ink if he can only put one ink in each pen? Answer: 0 Solution: He has options to fill his
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 information11.7 Maximum and Minimum Values
Arkansas Tech University MATH 2934: Calculus III Dr. Marcel B Finan 11.7 Maximum and Minimum Values Just like functions of a single variable, functions of several variables can have local and global extrema,
More informationCSE 312: Foundations of Computing II Quiz Section #2: Inclusion-Exclusion, Pigeonhole, Introduction to Probability
CSE 312: Foundations of Computing II Quiz Section #2: Inclusion-Exclusion, Pigeonhole, Introduction to Probability Review: Main Theorems and Concepts Binomial Theorem: Principle of Inclusion-Exclusion
More informationIntroduction to probability
Introduction to probability Suppose an experiment has a finite set X = {x 1,x 2,...,x n } of n possible outcomes. Each time the experiment is performed exactly one on the n outcomes happens. Assign each
More informationSection Summary. Permutations Combinations Combinatorial Proofs
Section 6.3 Section Summary Permutations Combinations Combinatorial Proofs Permutations Definition: A permutation of a set of distinct objects is an ordered arrangement of these objects. An ordered arrangement
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 informationMath 3338: Probability (Fall 2006)
Math 3338: Probability (Fall 2006) Jiwen He Section Number: 10853 http://math.uh.edu/ jiwenhe/math3338fall06.html Probability p.1/7 2.3 Counting Techniques (III) - Partitions Probability p.2/7 Partitioned
More informationLESSON 2: THE INCLUSION-EXCLUSION PRINCIPLE
LESSON 2: THE INCLUSION-EXCLUSION PRINCIPLE The inclusion-exclusion principle (also known as the sieve principle) is an extended version of the rule of the sum. It states that, for two (finite) sets, A
More informationCounting integral solutions
Thought exercise 2.2 20 Counting integral solutions Question: How many non-negative integer solutions are there of x 1 +x 2 +x 3 +x 4 = 10? Thought exercise 2.2 20 Counting integral solutions Question:
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 informationFunctions of several variables
Chapter 6 Functions of several variables 6.1 Limits and continuity Definition 6.1 (Euclidean distance). Given two points P (x 1, y 1 ) and Q(x, y ) on the plane, we define their distance by the formula
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 oom 3-044 Problem 1. An electronic toy displays a 4 4 grid
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 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 informationWeek 1. 1 What Is Combinatorics?
1 What Is Combinatorics? Week 1 The question that what is combinatorics is similar to the question that what is mathematics. If we say that mathematics is about the study of numbers and figures, then combinatorics
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 informationSolutions to the 2004 CMO written March 31, 2004
Solutions to the 004 CMO written March 31, 004 1. Find all ordered triples (x, y, z) of real numbers which satisfy the following system of equations: xy = z x y xz = y x z yz = x y z Solution 1 Subtracting
More informationBRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 2006 Senior Preliminary Round Problems & Solutions
BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 006 Senior Preliminary Round Problems & Solutions 1. Exactly 57.4574% of the people replied yes when asked if they used BLEU-OUT face cream. The fewest
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 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 informationStanford University CS261: Optimization Handout 9 Luca Trevisan February 1, 2011
Stanford University CS261: Optimization Handout 9 Luca Trevisan February 1, 2011 Lecture 9 In which we introduce the maximum flow problem. 1 Flows in Networks Today we start talking about the Maximum Flow
More informationThe Product Rule can be viewed as counting the number of elements in the Cartesian product of the finite sets
Chapter 6 - Counting 6.1 - The Basics of Counting Theorem 1 (The Product Rule). If every task in a set of k tasks must be done, where the first task can be done in n 1 ways, the second in n 2 ways, and
More informationMat 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 informationCS1800 Discrete Structures Fall 2016 Profs. Aslam, Gold, Ossowski, Pavlu, & Sprague 7 November, CS1800 Discrete Structures Midterm Version C
CS1800 Discrete Structures Fall 2016 Profs. Aslam, Gold, Ossowski, Pavlu, & Sprague 7 November, 2016 CS1800 Discrete Structures Midterm Version C Instructions: 1. The exam is closed book and closed notes.
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 informationThe probability set-up
CHAPTER 2 The probability set-up 2.1. Introduction and basic theory We will have a sample space, denoted S (sometimes Ω) that consists of all possible outcomes. For example, if we roll two dice, the sample
More informationMath 454 Summer 2005 Due Wednesday 7/13/05 Homework #2. Counting problems:
Homewor #2 Counting problems: 1 How many permutations of {1, 2, 3,..., 12} are there that don t begin with 2? Solution: (100%) I thin the easiest way is by subtracting off the bad permutations: 12! = total
More informationAssignment 2. Due: Monday Oct. 15, :59pm
Introduction To Discrete Math Due: Monday Oct. 15, 2012. 11:59pm Assignment 2 Instructor: Mohamed Omar Math 6a For all problems on assignments, you are allowed to use the textbook, class notes, and other
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 informationCIS 2033 Lecture 6, Spring 2017
CIS 2033 Lecture 6, Spring 2017 Instructor: David Dobor February 2, 2017 In this lecture, we introduce the basic principle of counting, use it to count subsets, permutations, combinations, and partitions,
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 informationSection 5.2 Graphs of the Sine and Cosine Functions
A Periodic Function and Its Period Section 5.2 Graphs of the Sine and Cosine Functions A nonconstant function f is said to be periodic if there is a number p > 0 such that f(x + p) = f(x) for all x in
More informationMultiple Choice Questions for Review
Review Questions Multiple Choice Questions for Review 1. Suppose there are 12 students, among whom are three students, M, B, C (a Math Major, a Biology Major, a Computer Science Major. We want to send
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 informationModular Arithmetic. Kieran Cooney - February 18, 2016
Modular Arithmetic Kieran Cooney - kieran.cooney@hotmail.com February 18, 2016 Sums and products in modular arithmetic Almost all of elementary number theory follows from one very basic theorem: Theorem.
More informationMinimal tilings of a unit square
arxiv:1607.00660v1 [math.mg] 3 Jul 2016 Minimal tilings of a unit square Iwan Praton Franklin & Marshall College Lancaster, PA 17604 Abstract Tile the unit square with n small squares. We determine the
More informationSec$on Summary. Permutations Combinations Combinatorial Proofs
Section 6.3 Sec$on Summary Permutations Combinations Combinatorial Proofs 2 Coun$ng ordered arrangements Ex: How many ways can we select 3 students from a group of 5 students to stand in line for a picture?
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 informationMath 42, Discrete Mathematics
c Fall 2018 last updated 10/29/2018 at 18:22:13 For use by students in this class only; all rights reserved. Note: some prose & some tables are taken directly from Kenneth R. Rosen, and Its Applications,
More informationPROOFS OF SOME BINOMIAL IDENTITIES USING THE METHOD OF LAST SQUARES
PROOFS OF SOME BINOMIAL IDENTITIES USING THE METHOD OF LAST SQUARES MARK SHATTUCK AND TAMÁS WALDHAUSER Abstract. We give combinatorial proofs for some identities involving binomial sums that have no closed
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 informationAlgorithmique appliquée Projet UNO
Algorithmique appliquée Projet UNO Paul Dorbec, Cyril Gavoille The aim of this project is to encode a program as efficient as possible to find the best sequence of cards that can be played by a single
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 informationThe probability set-up
CHAPTER The probability set-up.1. Introduction and basic theory We will have a sample space, denoted S sometimes Ω that consists of all possible outcomes. For example, if we roll two dice, the sample space
More informationRestricted Permutations Related to Fibonacci Numbers and k-generalized Fibonacci Numbers
Restricted Permutations Related to Fibonacci Numbers and k-generalized Fibonacci Numbers arxiv:math/0109219v1 [math.co] 27 Sep 2001 Eric S. Egge Department of Mathematics Gettysburg College 300 North Washington
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 informationWeek 3 Classical Probability, Part I
Week 3 Classical Probability, Part I Week 3 Objectives Proper understanding of common statistical practices such as confidence intervals and hypothesis testing requires some familiarity with probability
More informationThe Hex game and its mathematical side
The Hex game and its mathematical side Antonín Procházka Laboratoire de Mathématiques de Besançon Université Franche-Comté Lycée Jules Haag, 19 mars 2013 Brief history : HEX was invented in 1942
More information1.6 Congruence Modulo m
1.6 Congruence Modulo m 47 5. Let a, b 2 N and p be a prime. Prove for all natural numbers n 1, if p n (ab) and p - a, then p n b. 6. In the proof of Theorem 1.5.6 it was stated that if n is a prime number
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 informationHow to divide things fairly
MPRA Munich Personal RePEc Archive How to divide things fairly Steven Brams and D. Marc Kilgour and Christian Klamler New York University, Wilfrid Laurier University, University of Graz 6. September 2014
More informationLINEAR EQUATIONS IN TWO VARIABLES
LINEAR EQUATIONS IN TWO VARIABLES What You Should Learn Use slope to graph linear equations in two " variables. Find the slope of a line given two points on the line. Write linear equations in two variables.
More informationMath Exam 2 Review. NOTE: For reviews of the other sections on Exam 2, refer to the first page of WIR #4 and #5.
Math 166 Fall 2008 c Heather Ramsey Page 1 Math 166 - Exam 2 Review NOTE: For reviews of the other sections on Exam 2, refer to the first page of WIR #4 and #5. Section 3.2 - Measures of Central Tendency
More informationToday s Topics. Sometimes when counting a set, we count the same item more than once
Today s Topics Inclusion/exclusion principle The pigeonhole principle Sometimes when counting a set, we count the same item more than once For instance, if something can be done n 1 ways or n 2 ways, but
More informationMath Exam 2 Review. NOTE: For reviews of the other sections on Exam 2, refer to the first page of WIR #4 and #5.
Math 166 Fall 2008 c Heather Ramsey Page 1 Math 166 - Exam 2 Review NOTE: For reviews of the other sections on Exam 2, refer to the first page of WIR #4 and #5. Section 3.2 - Measures of Central Tendency
More information12th Bay Area Mathematical Olympiad
2th Bay Area Mathematical Olympiad February 2, 200 Problems (with Solutions) We write {a,b,c} for the set of three different positive integers a, b, and c. By choosing some or all of the numbers a, b and
More informationOutcome 7 Review. *Recall that -1 (-5) means
Outcome 7 Review Level 2 Determine the slope of a line that passes through A(3, -5) and B(-2, -1). Step 1: Remember that ordered pairs are in the form (x, y). Label the points so you can substitute into
More informationPermutation Groups. Every permutation can be written as a product of disjoint cycles. This factorization is unique up to the order of the factors.
Permutation Groups 5-9-2013 A permutation of a set X is a bijective function σ : X X The set of permutations S X of a set X forms a group under function composition The group of permutations of {1,2,,n}
More informationThe Art of Counting. Bijections, Double Counting. Peng Shi. September 16, Department of Mathematics Duke University
The Art of Counting Bijections, Double Counting Peng Shi Department of Mathematics Duke University September 16, 2009 What we focus on in this talk? Enumerative combinatorics is a huge branch of mathematics,
More informationSolutions for the Practice Final
Solutions for the Practice Final 1. Ian and Nai play the game of todo, where at each stage one of them flips a coin and then rolls a die. The person who played gets as many points as the number rolled
More informationn r for the number. (n r)!r!
Throughout we use both the notations ( ) n r and C n n! r for the number (n r)!r! 1 Ten points are distributed around a circle How many triangles have all three of their vertices in this 10-element set?
More informationFind the items on your list...but first find your list! Overview: Definitions: Setup:
Scavenger Hunt II A game for the piecepack by Brad Lackey. Version 1.1, 29 August 2006. Copyright (c) 2005, Brad Lackey. 4 Players, 60-80 Minutes. Equipment: eight distinct piecepack suits. Find the items
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 informationMath 1111 Math Exam Study Guide
Math 1111 Math Exam Study Guide The math exam will cover the mathematical concepts and techniques we ve explored this semester. The exam will not involve any codebreaking, although some questions on the
More information