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Question 1 (Module Outcome #1): A company wishes to offer employees a choice for their personal identification numbers (PIN). Each pin must be a strong of length 4. The first and second characters can be any digit 0, …, 9; the third and fourth character can be any of the 5 symbols chosen from {@,#,!,%,&}. No repeated characters are permitted. How many PINs are possible?

User Good Lux
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Answer:

1800

Explanation:

Since the first and second characters can be any digit 0, …, 9

So, there are 10 possible digits for first place

thus we have only 9 digits possible for second place

Also, it is given that the third and fourth character must be any of 5 given symbols

So, we have 5 possible symbols for third place

and since no repetition is allowed we have only 4 symbols left for the last place.

∴ Number of possible PINs are:

10 × 9 × 5 × 4 = 1800

User Emacs User
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