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If an employee ID code must consist of 3 non-repeating digits and each digit is in the code must be a prime number, how many ID codes can be generated?
(a) 4
(b) 10
(c) 22
(d) 24
(e) 26
source Jamboree
(a) 4
(b) 10
(c) 22
(d) 24
(e) 26
source Jamboree
28 Aug 2011, 03:35
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siddharthasingh
There are 4 prime number digits
2,3,5,7
We need to select 3 out of 4 and rearrange them in 3! ways.
So,
\(C^{4}_{3}*3!\)
OR, using permutation
\(P^{4}_{3}\)
Ans: 24.
***************************************
If the digits could be repeated:
\(4^3\)
29 Nov 2017, 22:05
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Hi All,
We're told that an ID code must consist of 3 non-repeating digits and each digit is in the code must be a PRIME number. We're asked for the number of ID codes that can be generated.
Since we're limited to just PRIME numbers/digits, we can only use the numbers 2, 3, 5 and 7. With a 3-digit code that uses NON-REPEATING digits, we can calculate the total number of possibilities by working through one digit at a time:
For the 1st digit, there are 4 options. Once we choose one....
For the 2nd digit, there are 3 options. Once we choose one...
For the 3rd digit, there are 2 options...
(4)(3)(2) = 24 possible ID codes.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
We're told that an ID code must consist of 3 non-repeating digits and each digit is in the code must be a PRIME number. We're asked for the number of ID codes that can be generated.
Since we're limited to just PRIME numbers/digits, we can only use the numbers 2, 3, 5 and 7. With a 3-digit code that uses NON-REPEATING digits, we can calculate the total number of possibilities by working through one digit at a time:
For the 1st digit, there are 4 options. Once we choose one....
For the 2nd digit, there are 3 options. Once we choose one...
For the 3rd digit, there are 2 options...
(4)(3)(2) = 24 possible ID codes.
Final Answer:
GMAT assassins aren't born, they're made,
Rich
