Precalculus (10th Edition)

$1600000.$
We know that the number of arrangements of $n$ objects in $r$ slots (where in a slot only $1$ of the $n$ elements can be put) is: $n^r$. If there are $k$ experiments, the first one can be done in $a_1$ ways, the second one in $a_2$ ways... the last one in $a_k$, then there are $a_1\cdot a_2\cdot...\cdot a_k$ ways of doing the experiments together. Hence here, since there are $8$ digits for the first choice, $2$ digits for the last choice, $10$ digits for the remaining five, the number of possibilities: $8\cdot10^5\cdot2=1600000.$