Probability of same numbers in a 5 digits password

I am trying to solve this probability problem but the result I get is wrong, would like some guidelines and tips.
Creating a random password made from 5 digits (0-9), need to calculate the probability that two numbers (for example 6 and 7) appear the same amount of times (could appear zero times as well).
The sample's size is $10^5$ (since it's a 5 digits password), and the wanted probability is as follows:
$5!$ for organizing the 5 numbers
$8^5$ in-case both numbers don't appear
${10\choose1}$${9\choose1}$${5\choose2}$ in-case for example 6 and 7 appeared once, the ${5\choose2}$ is for using the 6 and 7
${10\choose2}$${8\choose2}$${5\choose4}$ in-case each of those numbers was chosen two times
So,
$$\frac{5!(8^5+{10\choose1}{9\choose1}{5\choose2}+{10\choose2}{8\choose2}{5\choose4})} {10^5}$$
But the answer I get is bigger than 1 which is incorrect for probabilities...

The digits $6$ and $7$ appear twice each, choose the positions for the sixes, then for the sevens, then choose the other digit:

$$\binom{5}{2}\binom{3}{2}\times 8=240$$

The digits $6$ and $7$ appear once each, choose the position for the six, then for the seven, then choose the other digits:

$$\binom{5}{1}\binom{4}{1}\times 8^3=10240$$

The digits $6$ and $7$ do not appear:

$$8^5=32768$$

For a total of $43248$, giving the probability to be:

$$\frac{43248}{10^5}=0.43248$$