GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
All employees at Company W are assigned unique employee ID
[#permalink]
Updated on: 07 Dec 2022, 02:52
Updated on: 07 Dec 2022, 02:52
Expert Reply
1
Bookmarks
00:00
Question Stats:
57% (02:28) correct
42% (02:10) wrong based on 40 sessions
Hide Show timer Statistics
All employees at Company W are assigned unique employee ID codes that consist of numbers and letters from the alphabet. If letters can be repeated within the same code but numerals cannot, which of the following must be true?
Indicate all that apply
A. There are 2,600 possible different 3 digit codes consisting of 1 letter and 2 digits.
B. There are 60,840 possible different 4 digit codes consisting of 2 letters and 2 numbers.
C. There are 650 possible 2 digit codes consisting of 1 letter and 1 digit.
Indicate all that apply
A. There are 2,600 possible different 3 digit codes consisting of 1 letter and 2 digits.
B. There are 60,840 possible different 4 digit codes consisting of 2 letters and 2 numbers.
C. There are 650 possible 2 digit codes consisting of 1 letter and 1 digit.
Drill 2
Question: 12
Page: 565
Question: 12
Page: 565
Re: All employees at Company W are assigned unique employee ID
[#permalink]
12 Jan 2018, 15:35
12 Jan 2018, 15:35
1
Expert Reply
1
Bookmarks
Explanation
This is a permutation problem, so lay out your slots to fill and then multiply.
For choice (A) you have 1 slot for a letter and two slots for digits. There are 26 choices for your letter and 10 choices for your first digit. Because you can’t repeat digits, there are 9 choices for your second digit. Thus, the number of different codes that can be made is 26 × 10 × 9 = 2,340.Eliminate choice (A).
For correct choice (B), you have two spots for letters and two for digits, so you have 26 × 26 (letters can repeat) × 10 × 9 = 60,840.
Choice (C) is incorrect because the number of different codes consisting of one letter and one digit is 26 × 10 = 260. The only correct answer is choice (B).
This is a permutation problem, so lay out your slots to fill and then multiply.
For choice (A) you have 1 slot for a letter and two slots for digits. There are 26 choices for your letter and 10 choices for your first digit. Because you can’t repeat digits, there are 9 choices for your second digit. Thus, the number of different codes that can be made is 26 × 10 × 9 = 2,340.Eliminate choice (A).
For correct choice (B), you have two spots for letters and two for digits, so you have 26 × 26 (letters can repeat) × 10 × 9 = 60,840.
Choice (C) is incorrect because the number of different codes consisting of one letter and one digit is 26 × 10 = 260. The only correct answer is choice (B).
Re: All employees at Company W are assigned unique employee ID
[#permalink]
27 Feb 2020, 03:28
27 Feb 2020, 03:28
sandy wrote:
Explanation
This is a permutation problem, so lay out your slots to fill and then multiply.
For choice (A) you have 1 slot for a letter and two slots for digits. There are 26 choices for your letter and 10 choices for your first digit. Because you can’t repeat digits, there are 9 choices for your second digit. Thus, the number of different codes that can be made is 26 × 10 × 9 = 2,340.Eliminate choice (A).
For correct choice (B), you have two spots for letters and two for digits, so you have 26 × 26 (letters can repeat) × 10 × 9 = 60,840.
Choice (C) is incorrect because the number of different codes consisting of one letter and one digit is 26 × 10 = 260. The only correct answer is choice (B).
This is a permutation problem, so lay out your slots to fill and then multiply.
For choice (A) you have 1 slot for a letter and two slots for digits. There are 26 choices for your letter and 10 choices for your first digit. Because you can’t repeat digits, there are 9 choices for your second digit. Thus, the number of different codes that can be made is 26 × 10 × 9 = 2,340.Eliminate choice (A).
For correct choice (B), you have two spots for letters and two for digits, so you have 26 × 26 (letters can repeat) × 10 × 9 = 60,840.
Choice (C) is incorrect because the number of different codes consisting of one letter and one digit is 26 × 10 = 260. The only correct answer is choice (B).
Thank you for the explanation. But here, didn't we assume that the ID begins with letters and ends with digits ?
