Events & Promotions
|
|
GMAT 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.
Apr 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
25 Nov 2010, 18:49
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (01:45) correct
38%
(01:52)
wrong
based on 802
sessions
History
Date
Time
Result
Not Attempted Yet
A popular website requires users to create a password consisting of digits only. If no digit may be repeated and each password must be at least 9 digits long, how many passwords are possible?
A. 9! + 10!
B. 2 x 10!
C. 9! x 10!
D. 19!
E. 20!
A. 9! + 10!
B. 2 x 10!
C. 9! x 10!
D. 19!
E. 20!
26 Nov 2010, 10:26
Kudos
Bookmarks
shrive555
shrive555
Note that we are told that no digit may be repeated.
There are total of 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
As password must be at least 9 digits long then it can consist of either 9 or all 10 digits.
If password consists of 9 digits then # of passwords possible is \(C^9_{10}*9!=10*9!=10!\): \(C^9_{10}\) - # of ways to choose 9 digits which will be used in password and \(9!\) arranging these digits (or in another way \(P^9_{10}=10!\) - choosing 9 digits out of 10 when the order matters);
If password consists of all 10 digits then # of passwords possible is simply \(10!\);
So total # of passwords possible is \(10!+10!=2*10!\).
Answer: B.
25 Nov 2010, 19:48
Kudos
Bookmarks
If we choose the 10 different digits then they can be arranged (permutations) in 10! ways.
But the question asks at least 9 digits. So we have the possibility of choosing only 9 digits for the password ( but digit shouldn't repeat), so we can have a total of 10 different combinations and each combination can be arranged in 9! ways.
Therefore
10 x 9! + 10!
= 10! + 10!
= 2 x 10!
Therefore B
But the question asks at least 9 digits. So we have the possibility of choosing only 9 digits for the password ( but digit shouldn't repeat), so we can have a total of 10 different combinations and each combination can be arranged in 9! ways.
Therefore
10 x 9! + 10!
= 10! + 10!
= 2 x 10!
Therefore B
