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12 Apr 2021, 04:45
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The digits 0 to 9 are used to form three digit codes; however, there are a few conditions to be followed: the first digit cannot be 0 or 9, the second digit must be 0 or 9, and the second and third digits cannot both be ‘9’ in the same code. If the digits may be repeated in the same code, how many different codes are possible?
(A) 152
(B) 156
(C) 160
(D) 729
(E) 1000
(A) 152
(B) 156
(C) 160
(D) 729
(E) 1000
Thelegend2631
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Updated on: 19 Apr 2021, 07:07
Originally posted by Thelegend2631 on 12 Apr 2021, 04:57.
Last edited by Thelegend2631 on 19 Apr 2021, 07:07, edited 1 time in total.
Last edited by Thelegend2631 on 19 Apr 2021, 07:07, edited 1 time in total.
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Lets assign ABC for three places
A can have 8 values
Lets say b is 0, C can have 10 values
Total = 8*10 = 80
Similarly when B is 9 , c can have only 9 values since 9 is not possible
Total = 8*9 = 72
Total ways = 80+72 = 152
Answer A
Posted from my mobile device
A can have 8 values
Lets say b is 0, C can have 10 values
Total = 8*10 = 80
Similarly when B is 9 , c can have only 9 values since 9 is not possible
Total = 8*9 = 72
Total ways = 80+72 = 152
Answer A
Posted from my mobile device
19 Apr 2021, 06:47
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Hemanthdasu13
Lets assign ABC for three places
A can have 8 values
Lets say b is 0, C can have 10 values
Total = 8*10 = 80
Similarly when B is 9 , c can have only 8 values since 9 is not possible
Total = 8*9 = 72
Total ways = 80+72 = 152
Answer A
Posted from my mobile device
A can have 8 values
Lets say b is 0, C can have 10 values
Total = 8*10 = 80
Similarly when B is 9 , c can have only 8 values since 9 is not possible
Total = 8*9 = 72
Total ways = 80+72 = 152
Answer A
Posted from my mobile device
Hello, good explanation, thank you! I just noticed one small typo: When B is 9, C can have only 9 values (not 8).
