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25 Nov 2010, 18:49
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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%
(01:51)
wrong
based on 809
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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
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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
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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
