By László Lovász, József Pelikán, Katalin L. Vesztergombi

ISBN-10: 0387955844

ISBN-13: 9780387955841

Discrete arithmetic is readily turning into essentially the most vital parts of mathematical examine, with purposes to cryptography, linear programming, coding conception and the idea of computing. This e-book is geared toward undergraduate arithmetic and desktop technology scholars attracted to constructing a sense for what arithmetic is all approximately, the place arithmetic may be important, and what different types of questions mathematicians paintings on. The authors speak about a couple of chosen effects and strategies of discrete arithmetic, typically from the parts of combinatorics and graph thought, with a bit quantity conception, likelihood, and combinatorial geometry. anywhere attainable, the authors use proofs and challenge fixing to aid scholars comprehend the suggestions to difficulties. additionally, there are lots of examples, figures and routines unfold in the course of the ebook.

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

**Example text**

It is clear that we can determine who gets what by choosing the order in which the presents are laid out. There are n! ways to order the presents. But of course, the number n! overcounts the number of ways to distribute the presents, since many of these orderings lead to the same results (that is, every child gets the same set of presents). The question is, how many? So let us start with a given distribution of presents, and let’s ask the children to lay out the presents for us, nicely in a row, starting with the ﬁrst child, then continuing with the second, third, etc.

So we obtained 500 times 1001, which makes 500500. 2: 1000 · 1001/2 = 500500. 1) for the sum of the ﬁrst n odd numbers? 8 Prove that the sum of the ﬁrst n squares 1 + 4 + 9 + · · · + n2 is n(n + 1)(2n + 1)/6. 9 Prove that the sum of the ﬁrst n powers of 2 (starting with 1 = 20 ) is 2n − 1. ” instead of further repetition. There is nothing wrong with this, if the argument is suﬃciently simple so that we can intuitively see where the repetition leads. ” in cases where the outcome of the repetition is not so transparent.

Consider a set S with n elements, and ﬁx any element a ∈ S. We want to count the subsets of S. Let us divide them into two classes: those containing a and those not containing a. We count them separately. First, we deal with those subsets that don’t contain a. If we delete a from S, we are left with a set S with n − 1 elements, and the subsets we are interested in are exactly the subsets of S . By the induction hypothesis, the number of such subsets is 2n−1 . Second, we consider subsets containing a.

### Discrete Mathematics: Elementary and Beyond (Undergraduate Texts in Mathematics) by László Lovász, József Pelikán, Katalin L. Vesztergombi

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