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I am new to web development.

I am still on planning stage. One of the question that bug me is, lets say: the username can only contains 8 characters and each characters is not allowed special characters (@#$%^&), it only allows lower case "a-z", upper case "A-Z", and numerical number from "0-9". How many users I can have and how do u calculate it?

Case 1: Characters cannot be repeated eg "abcdefgh", "a1b2c3d4" etc

Case 2: Characters can be repeated, eg. "AAbbccdd", "a1b1c1D1" etc

FYI, the 8 just a sample number, of course the bigger number I provide, the better it is, but I am just curious and my math sucks :(

(I assume, need to use permutations??)

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up vote 7 down vote accepted

There are 26 uppercase letters, 26 lowercase letters, and 10 numerical digits, giving 62 possibilities for each of the eight characters. So there are $62\times 62\times \ldots \times 62 = 62^8$ possible usernames, which is a massive number.

This is in case 2, when anything goes. In case 1, once you have chosen a character, you are left with 61 possibilities for the next, then 60 possibilities for the next, giving the answer as $62\times 61\times \ldots \times 55$ usernames.

In general, therefore, with a string of $n$ characters you have $62^n$ possible usernames. With no repetitions, this becomes $\frac{62!}{(62-n)!}$ usernames.

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Your answer was quicker and has the general case as well, I will delete mine. I am glad, however, to see that I did not make any mistakes here. I usually do... :-) – Asaf Karagila May 7 '11 at 18:03
I understand case 2, pretty straight-forward: 62^8, but your case 1 abit confuse, are you saying the possibility are: 62*61*60*59*58*57*56*55?? Is it same as @Asaf answer? – FailMath May 7 '11 at 18:06
@Asaf, please dont delete your answer. You answer might be helpful to others users (I guess). But, as @Fahad answer first, and if its proven correct, I need to choose his answer as best answer, bcuz he is faster than you. Hope you dont mind. – FailMath May 7 '11 at 18:08
Yes that's it. You cannot use any character you've used before, so the number of possibilities goes down each time. I'm fairly sure Asaf wrote the same thing. – Fahad Sperinck May 7 '11 at 18:10
@FailMath: My answer was exactly as Fahad's only without regarding the general case. I don't see how two duplicate answers are useful here. :-) – Asaf Karagila May 7 '11 at 18:12

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