## Download Discrete Mathematics: Elementary and Beyond (Undergraduate by László Lovász, József Pelikán, Katalin L. Vesztergombi PDF

By László Lovász, József Pelikán, Katalin L. Vesztergombi

Discrete arithmetic is readily changing into the most vital components of mathematical examine, with functions to cryptography, linear programming, coding concept and the speculation of computing. This e-book is aimed toward undergraduate arithmetic and desktop technological know-how scholars attracted to constructing a sense for what arithmetic is all approximately, the place arithmetic might be worthy, and what types of questions mathematicians paintings on. The authors speak about a few chosen effects and techniques of discrete arithmetic, usually from the components of combinatorics and graph concept, with a bit quantity thought, likelihood, and combinatorial geometry. anyplace attainable, the authors use proofs and challenge fixing to aid scholars comprehend the strategies to difficulties. furthermore, there are lots of examples, figures and routines unfold in the course of the booklet.

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**Additional info for Discrete Mathematics: Elementary and Beyond (Undergraduate Texts in Mathematics)**

**Example text**

In how many ways can you send the postcards, if (a) there is a large number of each kind of postcard, and you want to send one card to each friend; (b) there is a large number of each kind of postcard, and you are willing to send one or more postcards to each friend (but no one should get two identical cards); (c) the shop has only 4 of each kind of postcard, and you want to send one card to each friend? 24 1. Let’s Count! 29 What is the number of ways to color n objects with 3 colors if every color must be used at least once?

2. 3) using the Binomial Theorem. 2): We n can replace n0 by n−1 (both are just 1), n1 by n−1 + n−1 0 0 1 , 2 by n−1 + n−1 1 2 , etc. 6 Identities in Pascal’s Triangle + · · · + (−1)n−1 n−1 n−1 + n−1 n−2 + (−1)n 51 n−1 , n−1 which is clearly 0, since the second term in each pair of brackets cancels with the ﬁrst term in the next pair of brackets. This method gives more than just a new proof of an identity we already know. What do we get if we start the same way, adding and subtracting binomial coeﬃcients alternatingly, but stop earlier?

Now, to get the total number of n-element subsets of S, we have to sum these numbers for all values of k = 0, 1, . . , n. 4). 3). (One could expect that, as with the “alternating” sum, we could get a nice formula for the sum obtained by stopping earlier, like n0 + n1 + · · · + nk . 4) is the coeﬃcient of xn y n in the expansion of (x + y)2n . Write (x + y)2n in the form (x + y)n (x + y)n , expand both factors (x + y)n using the Binomial Theorem, and then try to ﬁgure out the coeﬃcient of xn y n in the product.