DOCSLIB.ORG

3. Elementary Counting Problems 4.1,4.2. Binomial and Multinomial Theorems 2

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
3. Elementary Counting Problems 4.1,4.2. Binomial and Multinomial Theorems 2 3. Elementary Counting Problems 4.1,4.2. Binomial and Multinomial Theorems 2. Mathematical Induction Prof. Tesler Math 184A Fall 2017 Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 1 / 38 Multiplication rule Combinatorics is a branch of Mathematics that deals with systematic methods of counting things. Example How many outcomes (x, y, z) are possible, where x = roll of a 6-sided die; y = value of a coin flip; z = card drawn from a 52 card deck? (6 choices of x) (2 choices of y) (52 choices of z) = 624 × × Multiplication rule The number of sequences (x1, x2,..., xk) where there are n1 choices of x1, n2 choices of x2,..., nk choices of xk is n1 n2 nk. · ··· This assumes the number of choices of xi is a constant ni that doesn’t depend on the other choices. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 2 / 38 Cartesian product The Cartesian Product of sets A and B is A B = f (x, y): x A, y B g By the Multiplication× Rule, this has size2 jA 2Bj = jAj jBj. × · Example: f1, 2g f3, 4, 5g = f(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)g × The Cartesian product of sets A, B, and C is A B C = f (x, y, z): x A, y B, z C g This has size×jA ×B Cj = jAj jBj jC2j. 2 2 × × · · This extends to any number of sets. Example Roll of a 6-sided die A = f1, 2, 3, 4, 5, 6g jAj = 6 Value of a coin flip B = fH, Tg jBj = 2 Cards C = fA , 2 ,...g jCj = 52 ~ ~ The example on the previous slide becomes jA B Cj = 6 2 52 = 624. × × · · Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 3 / 38 Notation We often will need an n-element set. For n > 1, define [n] = f1, 2, 3,..., ng and also [0] = . ; Again, you may have seen [x] used for greatest integer, but we instead use x for floor and x for ceiling. b c d e Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 4 / 38 Powers Let A be a set. Ak = A A A (k times) × × · · · × jAkj = jAjk Example [2] = f1, 2g, with size j[2]j = j f1, 2g j = 2. [2]3 = f(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)g j[2]3j = 23 = 8 Example How many k letter strings are there over an n letter alphabet? 3-letter strings over the alphabet fa, b,..., zg: aaa, aab, aac,..., zzy, zzz There are 263 of them. In general, there are nk strings. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 5 / 38 Power set The power set of a set is the set of all of its subsets: P(S) = f A : A S g ⊆ P([3]) = P(f1, 2, 3g) = f , f1g , f2g , f3g , f1, 2g , f1, 3g , f2, 3g , f1, 2, 3gg ; Be careful on use of vs. : 2 ⊂ 1 < P([3]) P([3]) f1g P([3]) f1g , f2g , f1, 2g P([3]) ; 2 2 2 f g P([3]) ff1gg P([3]) ff1g , f2g , f1, 2gg P([3]) ; ⊂ ⊂ ⊂ How big is jP(S)j? Equivalently, how many subsets does a set have? Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 6 / 38 Number of subsets of an n-element set First solution How many subsets does an n element set have? We’ll use [n]; the solution will work for any set of size n, but it’s easier to work with a specific set. Make a sequence of decisions: Include 1 or not? Include 2 or not? Include··· n or not? Total: (2 choices)(2 choices) (2 choices) = 2n ··· Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 7 / 38 Number of subsets of an n-element set First solution Include 1? ; No Yes 1 Include 2? ; { } 2 1 1, 2 Include 3? ; { } { } { } ([3]) 3 2 2, 3 1 1, 3 1, 2 1, 2, 3 P ; { } { } { } { } { } { } { } Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 8 / 38 Number of subsets of an n-element set Second solution Consider a subset S [n]. ⊆ Form the word w1 wn or a sequence (w1,..., wn) ··· 1 if i S; wi = 2 0 otherwise. Example: The subset S = f1, 3, 4g of [5] is encoded as a word 10110 or as a sequence (1, 0, 1, 1, 0). Each subset of [n] gives a unique word in f0, 1gn and vice-versa. j f0, 1gn j = 2n, so there are 2n words and thus 2n subsets. This is called a bijective proof . Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 9 / 38 Function terminology Consider a function f : P Q. For element x in the set P, the function assigns a value f!(x) in the set Q. f is one-to-one iff for all x, y P, when f (x) = f (y) then x = y. 2 This is also called an injection. This means every element of Q either has exactly one inverse, or has no inverse. If f is one-to-one then jPj 6 jQj. f is onto iff for every z Q, there is at least one x P with f (x) = z. 2 2 This is also called a surjection. This means every element of Q has at least one inverse. If f is onto then jPj > jQj. f is a bijection iff it is one-to-one and onto. This means every element of Q has exactly one inverse. If f is a bijection then jPj = jQj. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 10 / 38 Function terminology f : P Q ! P Q For all functions, exactly one arrow leaves each element of P. One-to-one / Injection: At most one arrow hits each element of Q. Onto / Surjection: At least one arrow hits each element of Q. Bijection: Exactly one arrow hits each element of Q. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 11 / 38 Number of subsets of an n-element set Second solution, continued Define f : P([n]) f0, 1gn as follows: ! for S [n], form the word f (S) = w = w1 wn, where ⊆ ··· 1 if i S; wi = 2 0 otherwise. f is one-to-one: Suppose f (S) = f (T) = w. We need to show this requires S = T. Both S and T are subsets of [n]. For each i = 1,..., n, if wi = 1 then i S and i T, 2 2 while if wi = 0 then i < S and i < T. Thus, S and T have the exact same elements, so S = T. f is onto: Given w f0, 1gn, we must construct an inverse in the domain. There may2 be more than one inverse; we just have to construct one. S = f i [n]: wi = 1 g is in the domain and satisfies f (S) = w. 2 Thus, f is a bijection. So jP([n])j (the number of subsets of an n-element set) equals j f0, 1gn j = 2n. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 12 / 38 Number of subsets of an n-element set Third solution We will use Mathematical Induction (Chapter 2) to prove that the n number of subsets of [n] is 2 , for all integers n > 0: In general, the goal is to prove that a statement is true for all integers n > n0. Often, n0 is 0 or 1, but that’s not required. Base case: Show that the statement is true for n = n0. Sometimes it’s necessary to prove it specially for several other small values of n. Induction step: Assume that the statement holds for n. Use that to prove that it holds true for n + 1. Conclusion: the statement holds for all integers n > n0. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 13 / 38 Number of subsets of an n-element set Third solution n For all integers n > 0, the number of subsets of [n] is 2 . Base case First we show the statement is true for the smallest value of n (in this case, n = 0). When n = 0, [n] = [0] = has just one subset, which is . ; ; The formula gives 2n = 20 = 1. These agree, so the statement holds for the base case. Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 14 / 38 Number of subsets of an n-element set Third solution Induction step n For some n > 0, assume that the number of subsets of [n] is 2 . Use that to prove the number of subsets of [n + 1] is 2n+1 Split the subsets of [n + 1] into P Q, where [ P is the set of subsets of [n + 1] that don’t have n + 1, and Q is the set of subsets of [n + 1] that do have n + 1. P is the set of subsets of [n]. By the induction hypothesis, jPj = 2n. Insert n + 1 into each set in P to form Q. Thus, jPj = jQj. E.g., for n = 2: P = , f1g , f2g , f1, 2g ; Q = f3g , f1, 3g , f2, 3g , f1, 2, 3g The total number of subsets of [n + 1] is jPj + jQj = 2(2n) = 2 n+1. (This is an example of the Addition Rule, to be covered next.) n Conclusion: For all integers n > 0, the number of subsets of [n] is 2 . Prof. Tesler Elementary Counting Problems Math 184A / Fall 2017 15 / 38 Addition rule Count the number of days in a year, as follows.
Load more

Recommended publications
  • Multipermutations 1.1. Generating Functions
    CORE Metadata, citation and similar papers at core.ac.uk Provided by Elsevier - Publisher Connector JOURNALOF COMBINATORIALTHEORY, Series A 67, 44-71 (1994) The r- Multipermutations SEUNGKYUNG PARK Department of Mathematics, Yonsei University, Seoul 120-749, Korea Communicated by Gian-Carlo Rota Received October 1, 1992 In this paper we study permutations of the multiset {lr, 2r,...,nr}, which generalizes Gesse| and Stanley's work (J. Combin. Theory Ser. A 24 (1978), 24-33) on certain permutations on the multiset {12, 22..... n2}. Various formulas counting the permutations by inversions, descents, left-right minima, and major index are derived. © 1994 AcademicPress, Inc. 1. THE r-MULTIPERMUTATIONS 1.1. Generating Functions A multiset is defined as an ordered pair (S, f) where S is a set and f is a function from S to the set of nonnegative integers. If S = {ml ..... mr}, we write {m f(mD, .... m f(mr)r } for (S, f). Intuitively, a multiset is a set with possibly repeated elements. Let us begin by considering permutations of multisets (S, f), which are words in which each letter belongs to the set S and for each m e S the total number of appearances of m in the word is f(m). Thus 1223231 is a permutation of the multiset {12, 2 3, 32}. In this paper we consider a special set of permutations of the multiset [n](r) = {1 r, ..., nr}, which is defined as follows. DEFINITION 1.1. An r-multipermutation of the set {ml ..... ran} is a permutation al ""am of the multiset {m~ ....
    [Show full text]
  • 40 Chapter 6 the BINOMIAL THEOREM in Chapter 1 We Defined
    Chapter 6 THE BINOMIAL THEOREM n In Chapter 1 we defined as the coefficient of an-rbr in the expansion of (a + b)n, and r tabulated these coefficients in the arrangement of the Pascal Triangle: n Coefficients of (a + b)n 0 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1 5 1 5 10 10 5 1 6 1 6 15 20 15 6 1 ... n n We then observed that this array is bordered with 1's; that is, ' 1 and ' 1 for n = 0 n 0, 1, 2, ... We also noted that each number inside the border of 1's is the sum of the two closest numbers on the previous line. This property may be expressed in the form n n n%1 (1) % ' . r&1 r r This formula provides an efficient method of generating successive lines of the Pascal Triangle, but the method is not the best one if we want only the value of a single binomial coefficient for a 100 large n, such as . We therefore seek a more direct approach. 3 It is clear that the binomial coefficients in a diagonal adjacent to a diagonal of 1's are the 40 n numbers 1, 2, 3, ... ; that is, ' n. Now let us consider the ratios of binomial coefficients 1 to the previous ones on the same row. For n = 4, these ratios are: (2) 4/1, 6/4 = 3/2, 4/6 = 2/3, 1/4. For n = 5, they are (3) 5/1, 10/5 = 2, 10/10 = 1, 5/10 = 1/2, 1/5.
    [Show full text]
  • Q-Combinatorics
    q-combinatorics February 7, 2019 1 q-binomial coefficients n n Definition. For natural numbers n; k,the q-binomial coefficient k q (or just k if there no ambi- guity on q) is defined as the following rational function of the variable q: (qn − 1)(qn−1 − 1) ··· (q − 1) : (1.1) (qk − 1)(qk−1 − 1) ··· (q − 1) · (qn−k − 1)(qn−k−1 − 1) ··· (q − 1) For n = 0, k = 0, or n = k, we interpret the corresponding products 1, as the value of the empty product. If we exclude certain roots of unity from the domain of q, (1.1) is equal to (qn − 1)(qn−1 − 1) ··· (qn−k+1 − 1) (1.2) (qk − 1)(qk−1 − 1) ··· (q − 1) · 1 (qn−1 + ::: + q + 1) ··· (q2 + q + 1)(q + 1) · 1 = : (qk−1 + ::: + q + 1) ··· (q2 + q + 1)(q + 1) · 1(qn−k−1 + ::: + q + 1) ··· (q2 + q + 1)(q + 1) · 1 As the notation starts to be overwhelming, we introduce to q-analogue of the positive integer n, n−1 denoted by [n] or [n]q, as q + ::: + q + 1; and the q-analogue of the factorial of the positive integer n, denoted by [n]! or [n]q!, as [n]q! = [1]q · [2]q ··· [n]q. With this additional notation, we also can say n [n] ! = q ; (1.3) k q [k]q![n − k]q! when we avoid certain roots of unity with q. It is easy too see from (1.1) that n n = = 1 0 q n q and that for every 0 ≤ k ≤ n the symmetry rule n n = k q n − k q holds.
    [Show full text]
  • The Binomial Theorem and Combinatorial Proofs Shagnik Das
    The Binomial Theorem and Combinatorial Proofs Shagnik Das Introduction As we live our lives, we are faced with innumerable decisions on a daily basis1. Can I get away with wearing these clothes again before I have to wash them? Which frozen pizza should I have for dinner tonight? What excuse should I use for skipping the gym today? While it may at times seem tiring to be faced with all these choices on a regular basis, it is this exercise of free will that separates us from the machines we live with.2 This multitude of choices, then, provides some measure of our own vitality | the more options we have, the more alive we truly are. What, then, could be more important than being able to count how many options one actually has?4 As we have already seen during our first week of lectures, one of the main protagonists in these counting problems is the binomial coefficient, whose definition is given below. n Definition 1. Given non-negative integers n and k, the binomial coefficient k denotes the number of subsets of size k of a set of n elements. Equivalently, it is the number of unordered choices of k distinct elements from a set of n elements. However, the binomial coefficient leads a double life. Not only does it have the above definition, but also the formula below, which we proved in lecture. Proposition 2. For non-negative integers n and k, n nk n(n − 1)(n − 2) ::: (n − k + 1) n! = = = : k k! k! k!(n − k)! In this note, we will prove several more facts about this most fascinating of creatures.
    [Show full text]
  • Chapter 3.3, 4.1, 4.3. Binomial Coefficient Identities
    Chapter 3.3, 4.1, 4.3. Binomial Coefficient Identities Prof. Tesler Math 184A Winter 2019 Prof. Tesler Binomial Coefficient Identities Math 184A / Winter 2019 1 / 36 Table of binomial coefficients n k k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 k = 6 n = 0 1 0 0 0 0 0 0 n = 1 1 1 0 0 0 0 0 n = 2 1 2 1 0 0 0 0 n = 3 1 3 3 1 0 0 0 n = 4 1 4 6 4 1 0 0 n = 5 1 5 10 10 5 1 0 n = 6 1 6 15 20 15 6 1 n n! Compute a table of binomial coefficients using = . k k! (n - k)! We’ll look at several patterns. First, the nonzero entries of each row are symmetric; e.g., row n = 4 is 4 4 4 4 4 0 , 1 , 2 , 3 , 4 = 1, 4, 6, 4, 1 , n n which reads the same in reverse. Conjecture: k = n-k . Prof. Tesler Binomial Coefficient Identities Math 184A / Winter 2019 2 / 36 A binomial coefficient identity Theorem For nonegative integers k 6 n, n n n n = including = = 1 k n - k 0 n First proof: Expand using factorials: n n! n n! = = k k! (n - k)! n - k (n - k)! k! These are equal. Prof. Tesler Binomial Coefficient Identities Math 184A / Winter 2019 3 / 36 Theorem For nonegative integers k 6 n, n n n n = including = = 1 k n - k 0 n Second proof: A bijective proof. We’ll give a bijection between two sets, one counted by the left n n side, k , and the other by the right side, n-k .
    [Show full text]
  • Basic Combinatorics
    Basic Combinatorics Carl G. Wagner Department of Mathematics The University of Tennessee Knoxville, TN 37996-1300 Contents List of Figures iv List of Tables v 1 The Fibonacci Numbers From a Combinatorial Perspective 1 1.1 A Simple Counting Problem . 1 1.2 A Closed Form Expression for f(n) . 2 1.3 The Method of Generating Functions . 3 1.4 Approximation of f(n) . 4 2 Functions, Sequences, Words, and Distributions 5 2.1 Multisets and sets . 5 2.2 Functions . 6 2.3 Sequences and words . 7 2.4 Distributions . 7 2.5 The cardinality of a set . 8 2.6 The addition and multiplication rules . 9 2.7 Useful counting strategies . 11 2.8 The pigeonhole principle . 13 2.9 Functions with empty domain and/or codomain . 14 3 Subsets with Prescribed Cardinality 17 3.1 The power set of a set . 17 3.2 Binomial coefficients . 17 4 Sequences of Two Sorts of Things with Prescribed Frequency 23 4.1 A special sequence counting problem . 23 4.2 The binomial theorem . 24 4.3 Counting lattice paths in the plane . 26 5 Sequences of Integers with Prescribed Sum 28 5.1 Urn problems with indistinguishable balls . 28 5.2 The family of all compositions of n . 30 5.3 Upper bounds on the terms of sequences with prescribed sum . 31 i CONTENTS 6 Sequences of k Sorts of Things with Prescribed Frequency 33 6.1 Trinomial Coefficients . 33 6.2 The trinomial theorem . 35 6.3 Multinomial coefficients and the multinomial theorem . 37 7 Combinatorics and Probability 39 7.1 The Multinomial Distribution .
    [Show full text]
  • 8.6 the BINOMIAL THEOREM We Remake Nature by the Act of Discovery, in the Poem Or in the Theorem
    pg476 [V] G2 5-36058 / HCG / Cannon & Elich cr 11-30-95 MP1 476 Chapter 8 Discrete Mathematics: Functions on the Set of Natural Numbers (b) Based on your results for (a), guess the minimum number of handshakes. See Exercise 19, page 469. number of moves required if you start with an arbi- Show trary number of n disks. (Hint: Tohelpseeapat- n~n 2 1! tern,add1tothenumberofmovesforn51, 2, 3, f ~n! 5 for every positive integer n. 4.) 2 (c) A legend claims that monks in a remote monastery 56. Suppose n is an odd positive integer not divisible by 3. areworkingtomoveasetof64disks,andthatthe Show that n 2 2 1 is divisible by 24. (Hint: Consider the world will end when they complete their sacred three consecutive integers n 2 1, n, n 1 1. Explain task. Moving one disk per second without error, 24 why the product ~n 2 1!~n 1 1! must be divisible by 3 hours a day, 365 days a year, how long would it take andby8.) to move all 64 disks? 57. Finding Patterns If an 5 Ï24n 1 1, (a) write out 55. Number of Handshakes Suppose there are n people the first five terms of the sequence $an%. (b) What odd at a party and that each person shakes hands with every integers occur in $an%? (c) Explain why $an% contains all other person exactly once. Let f~n! denote the total primes greater than 3. (Hint: UseExercise56.) 8.6 THE BINOMIAL THEOREM We remake nature by the act of discovery, in the poem or in the theorem.
    [Show full text]
  • Discrete Math for Computer Science Students
    Discrete Math for Computer Science Students Cliff Stein Ken Bogart Scot Drysdale Dept. of Industrial Engineering Dept. of Mathematics Dept. of Computer Science and Operations Research Dartmouth College Dartmouth College Columbia University ii c Kenneth P. Bogart, Scot Drysdale, and Cliff Stein, 2004 Contents 1 Counting 1 1.1 Basic Counting . 1 The Sum Principle . 1 Abstraction . 2 Summing Consecutive Integers . 3 The Product Principle . 3 Two element subsets . 5 Important Concepts, Formulas, and Theorems . 6 Problems . 7 1.2 Counting Lists, Permutations, and Subsets. 9 Using the Sum and Product Principles . 9 Lists and functions . 10 The Bijection Principle . 12 k-element permutations of a set . 13 Counting subsets of a set . 13 Important Concepts, Formulas, and Theorems . 15 Problems . 16 1.3 Binomial Coefficients . 19 Pascal’s Triangle . 19 A proof using the Sum Principle . 20 The Binomial Theorem . 22 Labeling and trinomial coefficients . 23 Important Concepts, Formulas, and Theorems . 24 Problems . 25 1.4 Equivalence Relations and Counting (Optional) . 27 The Symmetry Principle . 27 iii iv CONTENTS Equivalence Relations . 28 The Quotient Principle . 29 Equivalence class counting . 30 Multisets . 31 The bookcase arrangement problem. 32 The number of k-element multisets of an n-element set. 33 Using the quotient principle to explain a quotient . 34 Important Concepts, Formulas, and Theorems . 34 Problems . 35 2 Cryptography and Number Theory 39 2.1 Cryptography and Modular Arithmetic . 39 Introduction to Cryptography . 39 Private Key Cryptography . 40 Public-key Cryptosystems . 42 Arithmetic modulo n .................................. 43 Cryptography using multiplication mod n ...................... 47 Important Concepts, Formulas, and Theorems . 48 Problems . 49 2.2 Inverses and GCDs .
    [Show full text]
  • Combinatorial Proofs of Fibonomial Identities 1
    COMBINATORIAL PROOFS OF FIBONOMIAL IDENTITIES ARTHUR T. BENJAMIN AND ELIZABETH REILAND Abstract. Fibonomial coefficients are defined like binomial coefficients, with integers re- placed by their respective Fibonacci numbers. For example, 10 = F10F9F8 . Remarkably, 3 F F3F2F1 n is always an integer. In 2010, Bruce Sagan and Carla Savage derived two very nice k F combinatorial interpretations of Fibonomial coefficients in terms of tilings created by lattice paths. We believe that these interpretations should lead to combinatorial proofs of Fibonomial identities. We provide a list of simple looking identities that are still in need of combinatorial proof. 1. Introduction What do you get when you cross Fibonacci numbers with binomial coefficients? Fibono- mial coefficients, of course! Fibonomial coefficients are defined like binomial coefficients, with integers replaced by their respective Fibonacci numbers. Specifically, for n ≥ k ≥ 1, n F F ··· F = n n−1 n−k+1 k F F1F2 ··· Fk For example, 10 = F10F9F8 = 55·34·21 = 19;635. Fibonomial coefficients resemble binomial 3 F F3F2F1 1·1·2 n coefficients in many ways. Analogous to the Pascal Triangle boundary conditions 1 = n and n n n n n = 1, we have 1 F = Fn and n F = 1. We also define 0 F = 1. n n−1 n−1 Since Fn = Fk+(n−k) = Fk+1Fn−k + FkFn−k−1, Pascal's recurrence k = k + k−1 has the following analog. Identity 1.1. For n ≥ 2, n n − 1 n − 1 = Fk+1 + Fn−k−1 : k F k F k − 1 F n As an immediate corollary, it follows that for all n ≥ k ≥ 1, k F is an integer.
    [Show full text]
  • Some Identities Involving Polynomial Coefficients
    Some identities involving polynomial coefficients Nour-Eddine Fahssi Department of Mathematics, FSTM, Hassan Second University of Casablanca, BP 146, Mohammedia, Morocco. [email protected] Abstract By polynomial (or extended binomial) coefficients, we mean the coefficients in the ex- pansion of integral powers, positive and negative, of the polynomial 1 + t + + tm; m 1 being a fixed integer. We will establish several identities and summation formulæ· · · parallel≥ to those of the usual binomial coefficients. 2010 AMS Classification Numbers: 11B65, 05A10, 05A19. 1 Introduction and preliminaries Definition 1.1. Let m 1 be an integer and let p (t)=1+ t + + tm. The coefficients ≥ m · · · defined by 1 k n n def [t ] (p (t)) , if k Z , n Z = m ∈ ≥ ∈ (1.1) k 0, if k Z≥ or k >mn whenever n Z>, !m ( 6∈ ∈ are called polynomial coefficients [10] or extended binomial coefficients. When the rows are indexed by n and the columns by k, the polynomial coefficients form an array called polynomial array of degree m and denoted by Tm. The sub-array corresponding to positive n is termed T+ extended Pascal triangle of degree m, or Pascal-De Moivre triangle, and denoted by m. For instance, the array T3 begins as shown in Table 1. arXiv:1507.07968v2 [math.NT] 30 Mar 2016 T+ The study of the coefficients of m is an old problem. Its history dates back at least to De Moivre [4, 11] and Euler [14]. In modern times, they have been studied in many papers, including several in The Fibonacci Quarterly. See [16] for a systematic treatise on the subject and an extensive bibliography.
    [Show full text]
  • History of Statistics 2. Origin of the Normal Curve – Abraham Demoivre
    History of Statistics 2. Origin of the Normal Curve – Abraham DeMoivre (1667- 1754) The normal curve is perhaps the most important probability graph in all of statistics. Its formula is shown here with a familiar picture. The “ e” in the formula is the irrational number we use as the base for natural logarithms. μ and σ are the mean and standard deviation of the curve. The person who first derived the formula, Abraham DeMoivre (1667- 1754), was solving a gambling problem whose solution depended on finding the sum of the terms of a binomial distribution. Later work, especially by Gauss about 1800, used the normal distribution to describe the pattern of random measurement error in observational data. Neither man used the name “normal curve.” That expression did not appear until the 1870s. The normal curve formula appears in mathematics as a limiting case of what would happen if you had an infinite number of data points. To prove mathematically that some theoretical distribution is actually normal you have to be familiar with the idea of limits, with adding up more and more smaller and smaller terms. This process is a fundamental component of calculus. So it’s not surprising that the formula first appeared at the same time the basic ideas of calculus were being developed in England and Europe in the late 17 th and early 18 th centuries. The normal curve formula first appeared in a paper by DeMoivre in 1733. He lived in England, having left France when he was about 20 years old. Many French Protestants, the Huguenots, left France when the King canceled the Edict of Nantes which had given them civil rights.
    [Show full text]
  • Enumerative Combinatorics 5: Q-Analogues
    Enumerative Combinatorics 5: q-analogues Peter J. Cameron Autumn 2013 In a sense, a q-analogue of a combinatorial formula is simply another formula involving a variable q which has the property that, as q ! 1, the second formula becomes the first. Of course there is more to it than that; some q-analogues are more important than others. What follows is nothing like a complete treatment; I will concentrate on a particularly important case, the Gaussian or q-binomial coefficients, which are, in the above sense, q-analogues of binomial coefficients. 5.1 Definition of Gaussian coefficients The Gaussian (or q-binomial) coefficient is defined for non-negative integers n and k as n (qn − 1)(qn−1 − 1) ··· (qn−k+1 − 1) = k k−1 : k q (q − 1)(q − 1) ··· (q − 1) In other words, in the formula for the binomial coefficient, we replace each factor r by qr − 1. Note that this is zero if k > n; so we may assume that k ≤ n. qr − 1 Now observe that lim = r. This follows from l'H^opital's rule: both q!1 q − 1 numerator and denominator tend to 0, and their derivatives are rqr−1 and 1, whose ratio tends to r. Alternatively, use the fact that qr − 1 = 1 + q + ··· + qr−1; q − 1 and we can now harmlessly substitute q = 1 in the right-hand side; each of the r terms becomes 1. 1 Hence if we replace each factor (qr − 1) in the definition of the Gaussian coefficient by (qr − 1)=(q − 1), then the factors (q − 1) in numerator and denominator cancel, so the expression is unchanged; and now it is clear that n n lim = : q!1 k q k 5.2 Interpretations Quantum calculus The letter q stands for \quantum", and the q-binomial coefficients do play a role in \quantum calculus" similar to that of the ordi- nary binomial coefficients in ordinary calculus.
    [Show full text]