Let $n$ be a fixed positive integer. Which is greater?[list=1]
[*]The number of $n$-tuples of integers whose largest value is $7$ and whose smallest value
is $0$; or
[*]The number of ordered triples $(A, B, C)$ that satisfy the following property: $A$, $B$, $C$
are subsets of $\{1, 2, 3, \dots , n\}$, and neither $C\subseteq A\cup B$, nor $B\subseteq A\cup C$.
[/list]
Your answer can be: $(1)$, $(2)$, the two counts are equal, or it depends on $n$.
1. **Counting the number of \( n \)-tuples of integers whose largest value is \( 7 \) and whose smallest value is \( 0 \):**

   - Each of the \( n \) digits in the tuple must be an integer in the range \([0, 7]\), giving \( 8 \) choices for each digit. Thus, there are \( 8^n \) ways to form the \( n \)-tuple.
   - We need to subtract the cases where there is no \( 7 \) or no \( 0 \). If there is no \( 7 \), we only have \( 7 \) choices for each digit, resulting in \( 7^n \) ways. Similarly, there are \( 7^n \) ways to have no \( 0 \) in the tuple.
   - By the Principle of Inclusion and Exclusion (PIE), we must add back the cases where there is neither \( 7 \) nor \( 0 \). This gives \( 6 \) choices for each digit, resulting in \( 6^n \) ways.
   - Therefore, the total count for (1) is:
     \[
     8^n - 2 \cdot 7^n + 6^n
     \]

2. **Counting the number of ordered triples \((A, B, C)\) that satisfy the given property:**

   - For each of the \( n \) elements in the set \(\{1, 2, \ldots, n\}\), each element can be part of \( A \) or not, giving \( 2^n \) possible sets for \( A \). Similarly, there are \( 2^n \) possible sets for \( B \) and \( 2^n \) possible sets for \( C \), resulting in \( 2^n \cdot 2^n \cdot 2^n = 8^n \) ways to have subsets \( A, B, C \).
   - We need to subtract the cases where \( C \subseteq A \cup B \) or \( B \subseteq A \cup C \).
   - To count the number of ways in which \( C \subseteq A \cup B \), we consider the cardinality of the set \( A \cup B \). Let this cardinality be \( k \). There are \(\binom{n}{k}\) possible sets of \( A \cup B \) with cardinality \( k \). The number of ways \( A \cup B \) can occur given cardinality \( k \) is:
     \[
     \sum_{m=0}^{k} \binom{k}{m} 2^m = 3^k
     \]
     by the binomial theorem.
   - The number of ways \( C \) can be chosen such that \( C \subseteq A \cup B \) is \( 2^k \). Considering all valid \( k \) and adding cases for each \( k \), the total number of ways to have \( C \subseteq A \cup B \) is:
     \[
     \sum_{k=0}^{n} \binom{n}{k} 2^k 3^k = \sum_{k=0}^{n} \binom{n}{k} 6^k = 7^n
     \]
     by the binomial theorem.
   - By symmetry, the number of ways \( B \subseteq A \cup C \) is also \( 7^n \). Hence, we must subtract \( 2 \cdot 7^n \) from our total count of \( 8^n \).
   - We need to add back the cases where both \( C \subseteq A \cup B \) and \( B \subseteq A \cup C \). The number of ways this happens is:
     \[
     \sum_{j=0}^{n} \binom{n}{j} 3^j 3^{n-j} = \sum_{j=0}^{n} \binom{n}{j} 3^n = 6^n
     \]
     by the binomial theorem.
   - Therefore, the total count for (2) is:
     \[
     8^n - 2 \cdot 7^n + 6^n
     \]

Since both counts are equal for all \( n \), the answer is:

The final answer is \(\boxed{\text{both counts are equal}}\)