5. For sets $A, B, C$ (not necessarily distinct), the union $A \cup B \cup C=\{1,2,3,4,5,6,7,8,9,10\}$, then under this condition, the number of possible triples $(A, B, C)$ is (write the answer in the form $a^{b}$).
5. $7^{10}$.

As shown in the figure, $A \cup B \cup C$ is divided into seven non-overlapping parts $X, Y, Z, W$, $U, V, T$ (where $Y = A \cap B \cap C, T = A - B - C, X = A \cap B - C$, etc.). Thus, each of the elements $1, 2, \cdots, 9, 10$ has 7 possible allocation places in $A \cup B \cup C$, so the total number of allocation schemes for these elements is $7^{10}$. Any two different allocation schemes will result in different ordered triples $(A, B, C)$ (since $X, Y, Z, W$, $U, V, T$ are uniquely determined by the ordered triple $(A, B, C)$), therefore, the number of ordered triples $(A, B, C)$ is equal to the total number of allocation schemes for $1, 2, \cdots, 9, 10$, which is $7^{10}$.