Here's a go at an answer. Test this against your own counting to check if we are understanding the question in the same way.
Let $n_{k}\left(j\right)$ be the number of $k$ digit numbers drawn from an alphabet of $j$ different digits and let $N_{k}\left(j\right)$ be the number with exactly $j$ different digits. For example:
$$\begin{array}{lcl}
n_{k}\left(1\right) &=& 1,\\
n_{16}\left(2\right) &=& 2^{16},\\
n_{16}\left(3\right) &=& 3^{16},\\
n_{3}\left(2\right) &=& 2^{2},\\
N_{2}\left(2\right) &=& 2,\\
N_{3}\left(2\right) &=& 6,\\
N_{k}\left(k\right) &=& k!,
\end{array}$$
etc. Clearly $n_{k}\left(j\right)=j^{k}$.
To make a $k$ digit number with exactly $j$ different digits then you must exclude all possibilities with fewer digits, i.e. $n_{k}\left(j-1\right),n_{k}\left(j-2\right),$ etc. Each of these is weighted by the number of possible subsets of $j-i$ digits from the full alphabet of $j$ digits. For example, in the case $j\in\left\{ 0,1\right\} , n_{2}\left(2\right)=4$ corresponds to the set $\left\{ 00,01,10,11\right\} $. But to get $N_{2}\left(2\right)$ you must exclude the two cases $\left\{ 00,11\right\} $. Thus
$$\begin{array}{lcl}
N_{k}\left(j\right) &=& n_{k}\left(j\right)+\sum_{i=1}^{j-1} (-1)^i \binom{j}{j-i}n_{k}\left(j-i\right) \\
&=& j^{k}+\sum_{i=1}^{j-1} (-1)^i \binom{j}{j-i}\left(j-i\right)^{k} \\
&=& \sum_{i=0}^{j} (-1)^i \binom{j}{j-i}\left(j-i\right)^{k}
\end{array}.$$
The factors of -1 account for the multiple counting of numbers in the sum. I'm not aware of a closed form expression for this sum. Mathematica won't give me one, but someone here would probably know if there is a simple expression for it. Note that when $j=k$ the series reduces to $k!$ as it should.
The answer to your question would be $\binom{10}{8} N_{16}(8) = 3886016697763200$.
EDIT: Note that this is related to Stirling numbers of the second kind:
$$ N_k(j) = j! S(k, j).$$
