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You will have 120 possibles (combinations) of passwords with 2 digits and 3 letters.

5! = 5 x 4 x 3 x 2 x 1 = 120

12abc

12acb

12bac

12bca

12cab

12cba

21abc

21acb

21bac

21bca

21cab

21cba

and so on up to 120 possibilities.

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Q: How many passwords are possibles if you have 2 digits and 3 letters?
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How many different 5 character passwords can you create from the following situations The code can be a mixture of letters and digits in any order and is case sensitive?

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