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.
 A: 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 $62\cdot61^{10-1}+62\cdot61^{11-1}+62\cdot61^{12-1}$

