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Joined: 16 May 2014
Posts: 592
Given Kudos: 0
GRE 1: Q165 V161
A popular website requires users
[#permalink]
Updated on: 28 Jun 2021, 20:12
Updated on: 28 Jun 2021, 20:12
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Question Stats:
50% (02:16) correct
50% (01:28) wrong based on 22 sessions
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A popular website requires users requires users to create passwords with digits only. The only rules is that no digit will be repeated and the passwords has to be at least 9 digits long. How many passwords are possible?
A. \(9!+10!\)
B. \(2*10!\)
C. \(9!*10!\)
D. \(19!\)
E. \(20!\)
Kudos to the best explanation! Truth is this one is a freebie in GRE.
A. \(9!+10!\)
B. \(2*10!\)
C. \(9!*10!\)
D. \(19!\)
E. \(20!\)
Kudos to the best explanation! Truth is this one is a freebie in GRE.
Originally posted by soumya1989 on 11 May 2016, 05:42.
Last edited by rx10 on 28 Jun 2021, 20:12, edited 3 times in total.
Last edited by rx10 on 28 Jun 2021, 20:12, edited 3 times in total.
Added OA and Tags
Re: A popular website requires users
[#permalink]
27 May 2016, 17:02
27 May 2016, 17:02
Without Repeatition of any digits, there will be two possible cases:
(1) At least 9 digits : 9! combinations
(2) At most 10 digits : 10! combinations
Total combination = 9! + 10!
Ans: A
(1) At least 9 digits : 9! combinations
(2) At most 10 digits : 10! combinations
Total combination = 9! + 10!
Ans: A
Re: A popular website requires users
[#permalink]
11 Jun 2016, 03:47
11 Jun 2016, 03:47
1
If the password has 9 digits then there are 10P9 way to arrange 10 digits in 9 places or 10!/(10-9)!= 10! ways
If the password has 10 digits the there are 10P10 way to arrange 10 digits in 10 places or 10!/(10-10)!= 10! ways
SO the answer is 10!+10! = 2*10!, Hence B
If the password has 10 digits the there are 10P10 way to arrange 10 digits in 10 places or 10!/(10-10)!= 10! ways
SO the answer is 10!+10! = 2*10!, Hence B
