/*! 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} Q37 Computer Variable Names A common... [FREE SOLUTION] | 91影视

91影视

Computer Variable Names A common computer programming rule is that names of variables must be between one and eight characters long. The first character can be any of the 26 letters, while successive characters can be any of the 26 letters or any of the 10 digits. For example, allowable variable names include A, BBB, and M3477K. How many different variable names are possible? (Ignore the difference between uppercase and lowercase letters)

Short Answer

Expert verified

The number of different variable names possible is equal to 2,095,681,645,538.

Step by step solution

01

Given information

A variable name can be between one and eight characters long. The first character should be one of the 26 letters, and the remaining characters can either be a letter from the 26 letters or any of the 10 digits.

02

State the counting rule

The number of possible ways a situation can take place is referred to as counts of arrangements for that event. The counting rule, permutation rule, and combination rule are a few counting techniques to find the counts.

03

Compute the counts for each length

Here, repetition is allowed.

As per the length of the variable, eight different cases are possible.

Case 1: The variable has one character.

The total number of letters to choose from is equal to 26.

The number of different one character long names possible = 26.

Case 2: The variable has two characters

The number of letters to choose from for the first character = 26

The total number of possibilities for the second character is the sum of the number of letters and the number of digits =26+10=36

The total number of two character long names is equal to:

2636=936

The same rules apply to the remaining six cases corresponding to 3, 4, 5, 6, 7, and 8 characters long variable names.

The total number of names possible if the name is three characters long is equal to

263636=33696

The total number of names possible if the name is four characters long is equal to

26363636=1213056

The total number of names possible if the name is five characters long is equal to

2636363636=43670016

The total number of names possible if the name is six characters long is equal to

263636363636=1572120576

The total number of names possible if the name is seven characters long is equal to

26363636363636=56596340736

The total number of names possible if the name is eight characters long is equal to

2636363636363636=2037468266496

Summarize the counts as follows:

Length of the variable

Number of ways to name

1

26

2

936

3

33,696

4

1,213,056

5

43,670,016

6

1,572,120,576

7

56,596,340,736

8

2,037,468,266,496

04

Compute the total counts of names

The total number of different names is the sum of the different names possible for each of the one to eight characters long names.

Length of the variable

Number of ways to name

1

26

2

936

3

33,696

4

1,213,056

5

43,670,016

6

1,572,120,576

7

56,596,340,736

8

2,037,468,266,496

Total

2,095,681,645,538

Thus, the total number of different variable names possible is equal to 2,095,681,645,538.

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 91影视!

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 Exercises 5鈥36, express all probabilities as fractions.

Sorting Hat At Hogwarts School of Witchcraft and Wizardry, the Sorting Hat chooses one of four houses for each first-year student. If 4 students are randomly selected from 16 available students (including Harry Potter), what is the probability that they are the four youngest students?

Odds. In Exercises 41鈥44, answer the given questions that involve odds.

Finding Odds in Roulette A roulette wheel has 38 slots. One slot is 0, another is 00, and the others are numbered 1 through 36, respectively. You place a bet that the outcome is an odd number.

a. What is your probability of winning?

b. What are the actual odds against winning?

c. When you bet that the outcome is an odd number, the payoff odds are 1:1. How much profit do you make if you bet \(18 and win?

d. How much profit would you make on the \)18 bet if you could somehow convince the casino to change its payoff odds so that they are the same as the actual odds against winning? (Recommendation: Don鈥檛 actually try to convince any casino of this; their sense of humor is remarkably absent when it comes to things of this sort.)

Odds. In Exercises 41鈥44, answer the given questions that involve odds.

Kentucky Pick 4 In the Kentucky Pick 4 lottery, you can place a 鈥渟traight鈥 bet of \(1 by selecting the exact order of four digits between 0 and 9 inclusive (with repetition allowed), so the probability of winning is 1/10,000. If the same four numbers are drawn in the same order, you collect \)5000, so your net profit is $4999.

a. Find the actual odds against winning.

b. Find the payoff odds.

c. The website www.kylottery.com indicates odds of 1:10,000 for this bet. Is that description accurate?

Oregon Pick 4 In the Oregon Pick 4 lottery game, a bettor selects four numbers between 0 and 9 and any selected number can be used more than once. Winning the top prize requires that the selected numbers match those and are drawn in the same order. Do calculations for this lottery involve the combinations rule or either of the two permutation rules presented in this section? Why or why not? If not, what rule does apply?

In Exercises 25鈥32, find the probability and answer the questions.

Genetics: Eye Color Each of two parents has the genotype brown/blue, which consists of the pair of alleles that determine eye color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one brown allele, that color will dominate and the eyes will be brown. (The actual determination of eye color is more complicated than that.)

a. List the different possible outcomes. Assume that these outcomes are equally likely.

b. What is the probability that a child of these parents will have the blue/blue genotype?

c. What is the probability that the child will have brown eyes?

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.