Enterbot3 Posted October 20, 2018 Report Share Posted October 20, 2018 Student passwords consist of three letters chosen from A to Z, followed by four digits chosen from 1–9. Repeated characters are allowed. How many possible passwords are there? [4 marks] Reply Link to post Share on other sites More sharing options...
Xx_Henry_xX Posted October 20, 2018 Report Share Posted October 20, 2018 Representing the three letters with A and the four digits with 1, some of the possible permutations are as follows: AAA1111 AA1A111 AA11A11 A1AA111 A1A1A11 There are 7!/(4!*3!) possible permutations in total. Each letter has 26 possibilities and each digit has 9 possibilities. (as 0 is not an option) This gives us 26^3 ways of choosing the three letters and 9^4 ways of choosing the four digits. (Order matters here: AAB ≠ ABA) Now we simply multiply the three results. That is, (26^3)*(9^4)*(7!/(4!*3!)) =17576*6561*35 =4,036,064,760 That easy. Reply Link to post Share on other sites More sharing options...
kw0573 Posted October 21, 2018 Report Share Posted October 21, 2018 @Xx_Henry_xX I believe letters "followed by" digits means precisely AAA1111, not any of the other options of mixing letters and numbers. So I think answer was just (26^3)(9^4). Possibly (52^3)(9^4) if including both capitals and lowercase. Reply Link to post Share on other sites More sharing options...
Xx_Henry_xX Posted October 21, 2018 Report Share Posted October 21, 2018 @kw0573 Could be. Guess I complicated stuffs by myself Reply Link to post Share on other sites More sharing options...
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