By Cliff L Stein, Robert Drysdale, Kenneth Bogart

Stein/Drysdale/Bogart's Discrete arithmetic for computing device Scientists is perfect for desktop technological know-how scholars taking the discrete math path. Written particularly for computing device technology scholars, this certain textbook at once addresses their wishes via delivering a starting place in discrete math whereas utilizing motivating, correct CS functions. this article takes an active-learning technique the place actions are provided as routines and the cloth is then fleshed out via factors and extensions of the workouts.

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**Extra resources for Discrete Mathematics for Computer Scientists **

**Example text**

2-2). However, each of these sets has two functions in it, one for each choice of fi (3). Thus, by the product principle, we have 4 · 2 = 8 functions from a three-element set to a two-element set. 2-2 are one-to-one, but f3 and f4 are not. A function f is called onto, or a surjection, if every element y in the range is f(x) for some x in the domain. 2-2 are onto, but f3 and f4 are not. 2-5 Using two- or three-element sets as domains and ranges, ﬁnd an example of a one-to-one function that is not onto.

Because there are k distinct elements of the list, every element of {1, 2, . . , k} appears in the list, so the function is onto. This means our function is a bijection. Thus, our deﬁnition of a permutation of a set is consistent with our deﬁnition of a k-element permutation in the case where the set is {1, 2, . . , k}. 5 In 16 Chapter 1: Counting permutations of {1, 2, 3, 4} are L = {123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241, 243, 312, 314, 321, 324, 341, 342, 412, 413, 421, 423, 431, 432}.

To write the problem algebraically, let Pi be the set of i-letter passwords and P be the set of all possible passwords. Clearly, P = P4 ∪ P5 ∪ P6 ∪ P7 ∪ P8 . The Pi are mutually disjoint; thus, we can apply the sum principle to obtain 8 |P | = |Pi |. i=4 We still need to compute |Pi |. For an i-letter password, there are 52 choices for the ﬁrst letter, 52 choices for the second, and so on. By the product principle, |Pi |—the number of passwords with i letters—is 52i . Therefore, the total number of passwords is 524 + 525 + 526 + 527 + 528 .