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A popular website requires users to create a password consis
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25 Nov 2010, 18:49
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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!
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shrive555 wrote: 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? 9! + 10! 2 x 10! 9! x 10! 19! 20! i got it right with slot method, but wasn't sure. shrive555 wrote: why 9! is multiplied with 10 ?
thanks 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.
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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



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why 9! is multiplied with 10 ? thanks
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Got it. 012345678 is one password 123456789 is another, by combination method we can find how many arrangements of 9 digits are possible, which is 10 Thanks B
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It's good to realize the permutation formula: n! / (nk)! We then got: 10! / (109)! = 10!/1! = 10! and 10! / (1010)! = 10!/0! = 10! = 10!/1! = 10! ( The answer is than: 2 * 10!
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Re: A popular website requires users to create a password consis
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15 May 2017, 21:39
Would have been brutal if 10! would have been one of the answers. It's easy to ignore the word atleast..



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Re: A popular website requires users to create a password consis
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12 Jun 2017, 01:40
shrive555 wrote: 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! The question must mention that the password can start with a '0'. Bunuel, if my reasoning is correct, would you please edit the question?
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Re: A popular website requires users to create a password consis
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12 Jun 2017, 02:43
ShashankDave wrote: shrive555 wrote: 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! The question must mention that the password can start with a '0'. Bunuel, if my reasoning is correct, would you please edit the question? A password/code can start with 0, it goes without saying.
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A popular website requires users to create a password consis
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12 Jun 2017, 09:56
9 digit passwords = \(10_C_9\) * 9! = 10! 10 digit passwords = \(10_C_{10}\) * 10!= 10! sum = 2*10! Answer B



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13 Jun 2017, 03:41
praneet87 wrote: Would have been brutal if 10! would have been one of the answers. It's easy to ignore the word atleast.. Couldn't agree more. Posted from my mobile device



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Re: A popular website requires users to create a password consis
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14 Jun 2017, 16:26
shrive555 wrote: 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! The number of ways to create a 10digit password from 10 digits is 10P10 = 10!. Thus, the number of ways to create the codes is 10! + 10! = 2(10!). Answer: B
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