# How many ways to make a password?

Suppose there are $62$ characters available for a password ($26$ uppercase letters, $26$ lowercase letters, and 10 numerals).

How many passwords are there that are exactly $12$ characters long if the passwords contain the same number of uppercase, lowercase, and numeral characters (without repetition)?

Why wouldn't the answer just be:

$(26×25×24×23)^2 + (10×9×8×7)$

Here's another: The password must be between 10 and 12 characters and may contain repetitions aside from the ith character being the same as the (i-1)th character or (i+1)th character. That is, the password cannot have immediate repetitions.

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I edited the answer for your second question. –  user31280 Nov 14 '12 at 14:50

You're to choose $4$ upper case, $4$ lower case and $4$ numeral characters. the total possible selection is $$12!\cdot \binom {26}{4}\cdot\binom {26}{4}\cdot\binom {10}{4}$$

Look at it this way

• There are $\binom {26}{4}$ ways of choosing $4$ uppercase letters from $26$
• There are $\binom {26}{4}$ ways of choosing $4$ lowercase letters from $26$
• There are $\binom {10}{4}$ ways of choosing $4$ numeral characters from $10$
• Therefore, there are $\binom {26}{4}\cdot\binom {26}{4}\cdot\binom {10}{4}$ possible ways of choosing $12$ characters in total.
• Considering that each character is distinct from the other. So we can permute each of our selection to form another password, thus we multiply by $12!$.

The second question can be simplified this way.

• The password can be $10$, $11$ or $12$ characters long.
• There is no fixed amount of each character type that the password must contain except that one can't have the same character side by side. So we're picking from a bowl of $62$ characters with replacement with our constraint.
• We have $62$ choices for the first character of our password, $61$ for the second (excluding the already chosen character), and $61$ for every following character of our password since we always exclude the immediate previously chosen character.
• It is more understandable this way. You pick a character from the bowl of $62$ and you write it down. You don't return the character because you don't want it to be the next one and then you pick a second character from the bowl of $61$, you write it down and you keep this second character(because you don't want it to be the third) while returning the first one, so you still have $61$ left in the bowl.
• Except the first choice which is $62$, you will always have $61$ choices.
• To choose $n$ character long password this way you have $62\cdot61^{n-1}$
• To choose $10$, $11$, or $12$, it is $62\cdot61^{10-1}+62\cdot61^{11-1}+62\cdot61^{12-1}$
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Can you elaborate more on the process? –  user1038665 Nov 14 '12 at 9:19
Particularly, where does the 12! come from? –  user1038665 Nov 14 '12 at 9:21
its that after you have chosen the 12 symbols for your password you can use them in any order –  Golob Nov 14 '12 at 9:23
@user1038665 You have $4$ choices of $26$ upper characters, $4$ choices of $26$ lower characters and $4$ choices of $10$ numerals as well as $12!$ possibilities to arrange them. –  k.stm Nov 14 '12 at 9:23
@user1038665: it is usually better to ask another question than to add a new one to an existing question. That way it is clearer what each answer refers to. –  Ross Millikan Nov 14 '12 at 14:58