Consider the problem "A common type of bike lock requires a six digit password, and only allows digits O through 9. How many passwords are there that contain the digit 9?" A student reasons: There are 6 possible places to put the 9, and then no restriction. So there are 105 possible arrangements of the other digits. This means the correct answer is 6 · 105. This solution is incorrect, since the student overcounts passwords containing more than one 1. This solution is incorrect, since, in the first step, the student needs to specify which digit gets the one. This solution is incorrect since there are not 10x10x10x10x10 arrangements of the digits O through 8 in the other three places. This solution is correct.

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter9: Counting And Probability
Section: Chapter Questions
Problem 3T: An Internet service provider requires its customers lo select a password consisting of four letter...
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Consider the problem "A common type of bike lock requires a six digit password, and only allows
digits O through 9. How many passwords are there that contain the digit 9?"
A student reasons: There are 6 possible places to put the 9, and then no restriction. So there are
105 possible arrangements of the other digits. This means the correct answer is 6 · 105.
This solution is incorrect, since the student overcounts passwords containing more than one 1.
This solution is incorrect, since, in the first step, the student needs to specify which digit gets the one.
This solution is incorrect since there are not 10x10x10x10x10 arrangements of the digits O through 8 in the
other three places.
This solution is correct.
Transcribed Image Text:Consider the problem "A common type of bike lock requires a six digit password, and only allows digits O through 9. How many passwords are there that contain the digit 9?" A student reasons: There are 6 possible places to put the 9, and then no restriction. So there are 105 possible arrangements of the other digits. This means the correct answer is 6 · 105. This solution is incorrect, since the student overcounts passwords containing more than one 1. This solution is incorrect, since, in the first step, the student needs to specify which digit gets the one. This solution is incorrect since there are not 10x10x10x10x10 arrangements of the digits O through 8 in the other three places. This solution is correct.
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