Given nine different numbers. Using each of them, how many different third-order determinants can we write down $?$
Let's first calculate how many ways we can rearrange the elements of a determinant without changing its value: Since

$$
\left|\begin{array}{lll}
A & B & C \\
D & E & F \\
G & H & I
\end{array}\right|=A E I+B F G+C D H-A F H-B D I-C E G
$$

we first need to decide whether to leave the $A E I$ group in place or to place it in the position of the $B F G$ or $C D H$ groups. This results in three different cases, and within each case, the new positions of the three elements can be chosen in 3! ways. The position of the $B F G$ group can be freely chosen from the remaining two positions, but the order of the elements is already determined, as they must belong to the negative groups containing the elements $I, A, E$ respectively. Similarly, the placement of the last three elements is uniquely determined, so in total there are $3 \cdot 3! \cdot 2 = 36$ arrangements with the same value. Since the 9 elements can be written into the determinant in $9!$ ways, the number of different values among these is generally $9! / 36 = 10080$. (With specific numbers, some of these values may be equal.)