Counting passwords

If I have an 8 digit passcode, taking values $0-9$, what is the probability that a randomly generated passcode has exactly $5$ occurrences of the same digit? What is the probability of having exactly one digit occur $4$ times in a passcode?

Solution

We have 10 digits to choose from, and 8 choose 5 choices of places. The remaining three digits can take any of the 9 remaining values, so there’s $9^3$ possibilities there. Divide it all by $9^8$. That is

\[\frac{10\binom{8}{5}9^3}{9^8}\]

For exactly 4 its a bit trickier, as we have some double counting. So its

\(\frac{10\binom{8}{4}9^4}{9^8} - \frac{10\binom{8}{4}9}{9^8}\)