# How would u solve this problem?

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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]

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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.

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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.

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Could be.

Guess I complicated stuffs by myself

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