Probability that your friend's password has no repeated characters 
You try to guess your friend's password, which is $8$ characters long, chosen from numbers the $0 \to 9$, letters a $\to$ z and letter A $\to $Z [caps]. But Your friend is lazy, and only choses from "a, s, d, f" and "A, S, D, F" and the number "$1$". What is the probability that your friend's password has no repeated characters? 

total possibilities: $9^8$
event possibilities: $9^{(8)}$ (9 permute 8)
$P = \frac{9^{(8)}}{9^8} - 0.008$?
But the answer the text gives is: $0.012$?
 A: A possibility to explain the answer $0.012$ could be that the password has to be alphanumeric, i.e.  it is not allowed to be all letters or all numbers. Under this restriction, the total number of possible passwords is 
given by $$9^8-8^8-1=   26.269.504$$
where $9^8$ is the total number of possibilities of filling $8 $places with $9$ characters, $8^8$ is the number of passwords containing only letters, and $1$ refers to the unique password containing only numbers $1$.
Considering then the passwords with no repeated characters, we can note that, to meet the condition of being alphanumeric, they all must contain the number $1$ one and only one time. As a result, they must also contain exactly $7$ letters. So their number is given by 
$$ \binom{8}{7} \cdot 7! \cdot 8=322.560$$
where $ \binom{8}{7} $ is the number of possibilities to choose $7$ letters from the group of available $8$ letters, $7! $ are the permutations of the chosen letters, and $8$ refers to the fact that the number  $1$ can be in any position within the password. 
Therefore, the resulting probability is
$$\frac {322.560}{26.269.504} \approx 0.012$$
A: There has to be a repetition because the set of possible characters is smaller than the password length.
This is called the Pigeonhole Principle
EDIT: This is outdated due to an edit to the original question, this was true for the character set "a","s","d","f","1" but not when "A","S","D","F" is added.
