/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 a. How many bit strings consist ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

a. How many bit strings consist of from one through four digits? (Strings of different lengths are considered distinct. Thus 10 and 0010 are distinct strings.) b. How many bit strings consist of from five through eight digits?

Short Answer

Expert verified
There are 30 bit strings consisting of 1 through 4 digits and 480 bit strings consisting of 5 through 8 digits.

Step by step solution

01

Part A: Bit strings with 1 to 4 digits

First let's find out how many bit strings there are for each length: - For 1 digit strings: There are 2 possibilities (0 or 1), so there are 2 bit strings. - For 2 digit strings: There are 2 possibilities for the first digit and 2 for the second digit, making a total of \(2 * 2 = 4\) bit strings. - For 3 digit strings: There are 2 possibilities for each of the three digits, making a total of \(2 * 2 * 2 = 8\) bit strings. - For 4 digit strings: There are 2 possibilities for the four digits, making a total of \(2 * 2 * 2 * 2 = 16\) bit strings. Now, we sum up the number of bit strings for each length: \(2 + 4 + 8 + 16 = 30\) So, there are 30 bit strings consisting of 1 through 4 digits.
02

Part B: Bit strings with 5 to 8 digits

Now let's find how many bit strings there are for bit string lengths of 5, 6, 7, and 8: - For 5 digit strings: There are 2 possibilities for each of the five digits, making a total of \(2 * 2 * 2 * 2 * 2 = 32\) bit strings. - For 6 digit strings: There are 2 possibilities for each of the six digits, making a total of \(2^6 = 64\) bit strings. - For 7 digit strings: There are 2 possibilities for each of the seven digits, making a total of \(2^7 = 128\) bit strings. - For 8 digit strings: There are 2 possibilities for each of the eight digits, making a total of \(2^8 = 256\) bit strings. Now, we sum up the number of bit strings for each length: \(32 + 64 + 128 + 256 = 480\) So, there are 480 bit strings consisting of 5 through 8 digits.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
At its core, combinatorics is a branch of mathematics focused on counting, arranging, and combining objects in certain orders or sets. This field is crucial for understanding the principles behind how many bit strings consist of a certain number of digits, as in the textbook exercise we’ve explored.

For example, when calculating bit strings of a certain length, combinatorics tells us we're looking into the number of ways we can arrange binary digits (0s and 1s). A string of length one has two possibilities because a binary digit can be either 0 or 1. For a string of length two, there are four possible combinations: 00, 01, 10, and 11. This pattern continues as the string increases in length. The key insight from combinatorics in this scenario is the way we calculate the total possibilities for each string length and then sum them to find the total number of distinct bit strings.
Binary Digits
Binary digits, also known as bits, are the fundamental units of digital data and can have a value of either 0 or 1. In the textbook exercise, we use bits to create bit strings, which are sequences of these binary digits.

In a computing context, bits serve as the smallest building block of data, representing a binary choice or state. This simplicity allows for binary systems to be stable and reliable, which is why they are the foundation of modern computing. When we consider bit strings of varying lengths, we start to see the implications of this binary system on the number of different sequences that can be formed, leading us into the concept of permutations.
Permutations
Permutations involve arranging a set of items in various orders where the arrangement is important. In a bit string context, each bit has two options, and as we create longer strings, we calculate permutations to identify all the possible combinations.

For instance, a 3-bit string has 2 options for each place, leading to a total of 2 x 2 x 2 permutations. Basically, for each additional binary digit we add to a string, we are effectively doubling the number of possible permutations. It’s this principle that allows us to swiftly calculate the number of bit strings for any given length and is a perfect illustration of the exponential growth inherent in such sequences.
Exponential Growth
Exponential growth occurs when the increase in a quantity is proportional to the current amount. This kind of growth is present in many natural phenomena and is critical for understanding the increase in the number of bit strings as their length grows.

In our bit string scenario, each additional digit results in an exponential increase of the possible combinations. The number of bit strings of a certain length is calculated by raising 2 (the number of possibilities for each digit) to the power of the string length. For instance, the exponential growth of bit strings with each additional digit is clear when you see that a 3-digit string has 8 possibilities (2^3) while a 4-digit string has 16 (2^4). Exponential growth helps highlight the substantial increase in complexity and quantity, even with the addition of a single binary digit to a string.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A non-leap year has 365 days. Assume that January 1 is a Monday. a. How many Sundays are there in the year? b. How many Mondays are there in the year?

If the largest of 56 consecutive integers is 279 , what is the smallest?

A bakery produces six different kinds of pastry. a. How many different selections of twenty pastries are there? b. Assuming that eclairs are one kind of pastry produced, how many different selections of twenty pastries are there if at least three must be eclairs? c. If a selection of twenty pastries is chosen randomly, what is the probability that at least three are eclairs? d. If a selection of twenty pastries is chosen randomly, what is the probability that exactly three are eclairs?

Suppose that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked for defects. a. How many different samples can be chosen? b. How many samples will contain at least one defective board? c. What is the probability that a randomly chosen sample of five contains at least one defective board?

An instructor gives an exam with twelve questions. Students are allowed to choose any ten to answer. a. How many different choices of ten questions are there? b. Suppose five questions require proof and seven do not. (i) How many groups of ten questions contain four that require proof and six that do not? (ii) How many groups of ten questions contain at least one that requires proof? (iii) How many groups of ten questions contain at most three that require proof? c. Suppose the exam instructions specify that at most one of questions 1 and 2 may be included among the ten. How many different choices of ten questions are there? d. Suppose the exam instructions specify that either both questions 1 and 2 are to be included among the ten or neither is to be included. How many different choices of ten questions are there?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.