Probability that a length $8$ password is made by $2$ digits and $6$ characters 
You have $10$ digits $(0,1,2,...,9)$ and $26$ characters (from the
  American alphabet).
Calculate the probability to automatic generate a 8-length password
  with $2$ digits and $6$ characters.

I know the answer is $ p=\frac{\binom{8}{2}\times10^2\times26^6}{36^8} $ and I know that $\binom{8}{2}$ are the picks where $2$ on $8$ elements are digits.
What I can't understand is why shouldn't I use $\binom{8}{6}$ to say that picks are made with $6$ characters on $8$ elements, instead of $\binom{8}{2}$?
The second question is: if I had to make a password with $1$ digits, $3$ special characters (chosen by $10$ special characters) and $4$ characters, the probability should change in: $ p=\frac{\binom{8}{1}\times\binom{8}{3}\times\binom{8}{4}\times10^1\times10^3\times26^4}{46^8} $?

You could use $8 \choose 6$ instead of $8 \choose 2$ just fine.  They are equal.  In general ${n \choose r}={n \choose n-r}$ and your proof is a good one.  For your second question, you should not use ${8 \choose 1}{8 \choose 3}{8 \choose 4}$.  The first $8 \choose 1$ picks which character will be special.  Having chosen that, you have only seven places left for the three digits, so the factor would be $7 \choose 3$.  Having chosen the first two, you have only four places left for four letters, so the overall factor should be ${8 \choose 1}{7 \choose 3}{4 \choose 4}=\frac {8!}{1!3!4!}$
