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A company ordered for security codes to be formed for each o
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04 Jan 2022, 11:37
04 Jan 2022, 11:37
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A company ordered for security codes to be formed for each of its employees. Codes should be designed such that each code has five characters comprising of 3 alphabets and 2 digits. If only 3 digits (1, 2, 3) can be used and only 2 alphabets (X, Y) can be used for the codes, then how many different codes can be formed such that repetition of the alphabets and digits is allowed?
A. 72
B. 180
C. 360
D. 720
E. 1440
Posted from my mobile device
A. 72
B. 180
C. 360
D. 720
E. 1440
Posted from my mobile device
Re: A company ordered for security codes to be formed for each o
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04 Jan 2022, 11:45
04 Jan 2022, 11:45
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Expert Reply
Each code has 5 characters
Each code will have 3 of these characters as letters (X or Y) and 2 of these characters as Digits (1 or 2 or 3)
(1st) out of the 5 character order, choose which 3 spaces the Letters will take and which 2 spaces the Digits will take
For instance we can have:
Letter - letter - letter - digit - digit
Or
Lettter - letter - digit - letter - digit
Etc.
The number of different unique arrangements is given by: 5! / 3! 2! = 10
AND
(2nd) for each one of these 10 arrangements, we need to choose from the available options of Digits and Characters. Repeatition is allowed.
Digit 1 - 3 available options
Digit 2 - 3 available options
Letter 1 - 2 available options
Letter 2 - 2 available options
Letter 3 - 2 available options
Applying the Factor Foundation Rule, the number of unique codes that can be formed is:
(10) * (3 * 3 * 2 * 2 * 2) =
(10) (9) (8) = 720 possible codes
Each code will have 3 of these characters as letters (X or Y) and 2 of these characters as Digits (1 or 2 or 3)
(1st) out of the 5 character order, choose which 3 spaces the Letters will take and which 2 spaces the Digits will take
For instance we can have:
Letter - letter - letter - digit - digit
Or
Lettter - letter - digit - letter - digit
Etc.
The number of different unique arrangements is given by: 5! / 3! 2! = 10
AND
(2nd) for each one of these 10 arrangements, we need to choose from the available options of Digits and Characters. Repeatition is allowed.
Digit 1 - 3 available options
Digit 2 - 3 available options
Letter 1 - 2 available options
Letter 2 - 2 available options
Letter 3 - 2 available options
Applying the Factor Foundation Rule, the number of unique codes that can be formed is:
(10) * (3 * 3 * 2 * 2 * 2) =
(10) (9) (8) = 720 possible codes
Re: A company ordered for security codes to be formed for each o
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11 Jul 2024, 09:05
11 Jul 2024, 09:05
Hello from the GRE Prep Club BumpBot!
Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
