By Barry Lewis, Alan Camina

Written for college kids taking a moment or 3rd yr undergraduate path in arithmetic or computing device technological know-how, this publication is the proper better half to a direction in enumeration. Enumeration is a department of combinatorics the place the elemental material is a number of equipment of trend formation and counting. An advent to Enumeration presents a finished and functional creation to this topic giving a transparent account of basic effects and a radical grounding within the use of strong recommendations and tools.

Two significant topics run in parallel in the course of the e-book, producing features and crew idea. the previous subject takes enumerative sequences after which makes use of analytic instruments to find how they're made up. team idea offers a concise creation to teams and illustrates how the idea can be utilized to count number the variety of symmetries a specific item has. those enhance and expand easy workforce rules and techniques.

The authors current their fabric via examples which are rigorously selected to set up key leads to a average environment. the purpose is to steadily construct primary theorems and strategies. This improvement is interspersed with workouts that consolidate rules and construct self belief. a few routines are associated with specific sections whereas others diversity throughout an entire bankruptcy. all through, there's an try and current key enumerative rules in a image manner, utilizing diagrams to cause them to instantly available. the improvement assumes a few simple team concept, a familiarity with analytic features and their energy sequence growth in addition to a few simple linear algebra.

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**Extra info for An Introduction to Enumeration (Springer Undergraduate Mathematics Series)**

**Sample text**

26 The recurrence relation satisfied by the Fibonacci sequence {Fr } is Fr = Fr−1 + Fr−2 with the initial values F0 = 0 and F1 = 1. Applying the three-step recipe gives: (i) The recurrence is Fr = Fr−1 + Fr−2 . (ii) Then we sum on the index r with corresponding powers of z: ∑ Fr zr = ∑ Fr−1 zr + ∑ Fr−2 zr . r 2 r 2 r 2 (iii) Finally ∑ Fr zr − F0 − F1 z = z ∑ Fr−1 zr−1 − F0 r 0 r 0 + z2 ∑ Fr zr . 3 Sequence to Generating Function 37 Using the initial values and denoting the generating function by F(z), we have F(z) − 0 − z = z (F(z) − 0) + z2 F(z) and we can solve this for the required generating function F(z): F(z) = z ∑ Fr zr = 1 − z − z2 .

1 − 3z + z2 Find an explicit form for the terms of the sequence. What is the connection between this sequence and the Fibonacci sequence? 2 It’s the Denominator that Counts There is another use of generating functions that we have not yet looked at. A generating function may be used to find a recurrence relation for the sequence it represents. 10 Given the generating function for a sequence {ur }: 5 − 13z ∑ ur zr = 1 − 5z + 6z2 r 0 we take the denominator over to the other side 1 − 5z + 6z2 ∑ ur zr = 5 − 13z.

26 The recurrence relation satisfied by the Fibonacci sequence {Fr } is Fr = Fr−1 + Fr−2 with the initial values F0 = 0 and F1 = 1. Applying the three-step recipe gives: (i) The recurrence is Fr = Fr−1 + Fr−2 . (ii) Then we sum on the index r with corresponding powers of z: ∑ Fr zr = ∑ Fr−1 zr + ∑ Fr−2 zr . r 2 r 2 r 2 (iii) Finally ∑ Fr zr − F0 − F1 z = z ∑ Fr−1 zr−1 − F0 r 0 r 0 + z2 ∑ Fr zr . 3 Sequence to Generating Function 37 Using the initial values and denoting the generating function by F(z), we have F(z) − 0 − z = z (F(z) − 0) + z2 F(z) and we can solve this for the required generating function F(z): F(z) = z ∑ Fr zr = 1 − z − z2 .