/*! 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 7 Recall, by 8-bit strings, we mea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Recall, by 8-bit strings, we mean strings of binary digits, of length 8 . (a) How many 8-bit strings are there total? (b) How many 8-bit strings have weight \(5 ?\) (c) How many subsets of the set \(\\{a, b, c, d, e, f, g, h\\}\) contain exactly 5 elements? (d) Explain why your answers to parts (b) and (c) are the same. Why are these questions equivalent?

Short Answer

Expert verified
There are 2^8 total 8-bit strings, 8 choose 5 (or C(8,5)) 8-bit strings with weight 5, and 8 choose 5 subsets of a set with 8 elements containing exactly 5 elements. Answers to (b) and (c) are the same because they both represent the number of ways to choose 5 elements from a set of 8.

Step by step solution

01

Counting Total 8-bit Strings

Each position in an 8-bit string can be either '0' or '1'. Therefore, for each of the 8 positions, there are 2 possible choices, leading to a total of 2^8 different 8-bit strings.
02

Counting 8-bit Strings with Weight 5

The weight of an 8-bit string is the number of '1's it contains. To have a weight of 5, exactly 5 out of the 8 positions must be '1's. This is a combination problem where we choose 5 positions from 8 to be '1', which is calculated by 8 choose 5 or C(8,5).
03

Counting Subsets with 5 Elements

To count the number of subsets containing exactly 5 elements from a set of 8 elements, we use combinations. We are selecting 5 elements from the set, which is again calculated by 8 choose 5 or C(8,5).
04

Explaining Equivalence

The problems in parts (b) and (c) are equivalent because they both involve selecting 5 elements out of an 8-element set. In (b), '1's in an 8-bit string represent selected elements, while in (c), the elements themselves are selected to be in the subset.

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.

Understanding Binary Digits
Binary digits, often referred to as bits, are the fundamental units of data in digital computer systems and communications. A binary digit can be either a 0 or a 1. When we talk about an 8-bit string, we are referring to a sequence of 8 binary digits. For example, 11001001 and 10010011 are two distinct 8-bit strings.

In understanding 8-bit strings, we can visualize each bit as an individual switch that has two states: on or off, or in binary terms, 1 or 0. This binary numbering system is base-2, unlike our common base-10 system, which has ten possible digits (0 to 9) for each place value. In the realm of computing and digital representation, binary digits hold significant importance due to their ease of implementation in electronic circuitry through logic gates and their role in representing any form of data.

Considering an 8-bit string, since each bit can be independently set to 0 or 1, we quickly realize that we can represent 256 unique combinations, from 00000000 to 11111111. This is calculated using exponentiation in base-2 (i.e., 2 raised to the power of 8, or 2^8).
Combinatorics in Action
Combinatorics is the field of mathematics concerned with counting, combination, and permutation of sets of elements. When we discuss the weight of an 8-bit string, or how many subsets of a given set contain a specific number of elements, we are delving into combinatorial mathematics.

For example, determining the number of 8-bit strings with a weight of 5 means finding all possible strings where exactly five bits are turned 'on' (or set to '1'). This requires us to use a specific combinatorial formula called the binomial coefficient, often represented as 'n choose k.' The notation C(n, k) or sometimes nCk is used, which expresses the number of ways to choose a subset of k elements from a set of n distinct elements without regard to the order of selection.

To find the answer for our 8-bit strings with a weight of 5, we calculate C(8, 5)—the number of ways to choose five positions from eight to be '1', ignoring all remaining that will be '0'. In general, the formula for the binomial coefficient is the factorial of n divided by the product of the factorial of k and the factorial of n minus k (n! / [k!(n - k)!]). Using this, we gain insight into various combinatorial problems.
Set Theory Applications
Set theory is a branch of mathematical logic that deals with sets, which are collections of objects. Counting subsets of a particular size from a larger set resembles the previous combinatorial problem. Set theory often encounters problems that involve understanding the nature of collections and how they can be combined, intersected, or disjointed.

In the case of counting the subsets consisting of 5 elements from the 8-element set {a, b, c, d, e, f, g, h}, we again deal with 'n choose k'—in this case, 8 choose 5. This scenario resembles the prior task of finding 8-bit strings with a weight of 5 because both involve selecting a limited number of elements from a larger set.

Understanding the parallels between binary strings and set theory enriches our comprehension of how these mathematical principles intersect—whether we are flipping bits within a digital string or choosing elements from a set, the fundamental combinatorial concepts remain constant. This deepens our appreciation for different mathematical domains' interconnectedness, particularly when reasoning about data structures in computer science.

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

In an attempt to clean up your room, you have purchased a new floating shelf to put some of your 17 books you have stacked in a corner. These books are all by different authors. The new book shelf is large enough to hold 10 of the books. (a) How many ways can you select and arrange 10 of the 17 books on the shelf? Notice that here we will allow the books to end up in any order. Explain. (b) How many ways can you arrange 10 of the 17 books on the shelf if you insist they must be arranged alphabetically by author? Explain.

Consider five digit numbers \(\alpha=a_{1} a_{2} a_{3} a_{4} a_{5},\) with each digit from the set \\{1,2,3,4\\} (a) How many such numbers are there? (b) How many such numbers are there for which the sum of the digits is even? (c) How many such numbers contain more even digits than odd digits?

For your college interview, you must wear a tie. You own 3 regular (boring) ties and 5 (cool) bow ties. (a) How many choices do you have for your neck-wear? (b) You realize that the interview is for clown college, so you should probably wear both a regular tie and a bow tie. How many choices do you have now? (c) For the rest of your outfit, you have 5 shirts, 4 skirts, 3 pants, and 7 dresses. You want to select either a shirt to wear with a skirt or pants, or just a dress. How many outfits do you have to choose from?

For how many three digit numbers \((100\) to 999\()\) is the sum of the digits even? (For example, 343 has an even sum of digits: \(3+4+3=10\) which is even.) Find the answer and explain why it is correct in at least two different ways.

The number 735000 factors as \(2^{3} \cdot 3 \cdot 5^{4} \cdot 7^{2}\). How many divisors does it have? Explain your answer using the multiplicative principle.

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.