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21 Sep 2020, 01:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
22% (02:17) correct
78%
(02:38)
wrong
based on 114
sessions
History
Date
Time
Result
Not Attempted Yet
A locker code consists of five digits, for example 04391, 23335, 05531... David wants to set his locker code so that digit 5 is not immediately followed by the digit 9. How many such codes are possible?
A. 96,000
B. 96,001
C. 96,003
D. 96,030
E. 96,040
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A. 96,000
B. 96,001
C. 96,003
D. 96,030
E. 96,040
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21 Sep 2020, 03:20
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Total number of arrangements = 10^5
Invalid arrangements =
5 9 _ _ _ : 1000
_ 5 9 _ _ : 1000
_ _ 5 9 _ : 1000
_ _ _ 5 9 : 1000
So, valid cases = 10^5 - 4000 = 96000, but there are 2 cases which were considered twice,
5 9 5 9 _ was considered once in 5 9 _ _ _ and once in _ _ 5 9 _, so we need to consider such cases only once. = 10
_ 5 9 5 9 was considered once in _ 5 9 _ _ and once in _ _ _ 5 9, so we need to consider such cases only once. = 10
5 9 _ 5 9 was considered once in 5 9 _ _ _ and once in _ _ _ 5 9, so we need to consider such cases only once. = 10
So the total number of cases = 10^5 - 4000 + 30 = 96030 option D
Invalid arrangements =
5 9 _ _ _ : 1000
_ 5 9 _ _ : 1000
_ _ 5 9 _ : 1000
_ _ _ 5 9 : 1000
So, valid cases = 10^5 - 4000 = 96000, but there are 2 cases which were considered twice,
5 9 5 9 _ was considered once in 5 9 _ _ _ and once in _ _ 5 9 _, so we need to consider such cases only once. = 10
_ 5 9 5 9 was considered once in _ 5 9 _ _ and once in _ _ _ 5 9, so we need to consider such cases only once. = 10
5 9 _ 5 9 was considered once in 5 9 _ _ _ and once in _ _ _ 5 9, so we need to consider such cases only once. = 10
So the total number of cases = 10^5 - 4000 + 30 = 96030 option D
General Discussion
Updated on: 21 Sep 2020, 03:12
Originally posted by Archit3110 on 21 Sep 2020, 01:05.
Last edited by Archit3110 on 21 Sep 2020, 03:12, edited 1 time in total.
Last edited by Archit3110 on 21 Sep 2020, 03:12, edited 1 time in total.
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Bunuel
since code can be repetitive so total ways to get 5 digit code 10^5 and possible ways where 5 is followed by 9 be X
X_l_l_l_l
X*10*10*10 ; which can be arranged in 4 ways or say 4*10^3
so digit 5 is not immediately followed by the digit 9
also possible 3 cases where we have repetition of given digits (9,5) and as per condition of David requires 599a5 where a is other digit total ways to arrange 5!/2!*2! ; 30
10^5-(4*10^3)+30 ; 96,030
option D
