Let $A=\{1,2,\ldots,n\}$. Find the number of unordered triples $(X,Y,Z)$ that satisfy $X\bigcup Y \bigcup Z=A$
To find the number of unordered triples \((X, Y, Z)\) that satisfy \(X \cup Y \cup Z = A\) where \(A = \{1, 2, \ldots, n\}\), we need to consider the following steps:

1. **Counting the ways to assign each element to subsets \(X\), \(Y\), and \(Z\):**
   Each element in \(A\) can be in any of the subsets \(X\), \(Y\), or \(Z\), or in any combination of these subsets. There are \(2^3 = 8\) possible ways to assign each element to the subsets \(X\), \(Y\), and \(Z\) (since each element can either be in or out of each subset).

2. **Excluding the empty assignment:**
   Out of these 8 possibilities, one possibility is that the element is not in any of the subsets \(X\), \(Y\), or \(Z\). This is not allowed since \(X \cup Y \cup Z = A\) must cover all elements of \(A\). Therefore, we have \(8 - 1 = 7\) valid ways to assign each element to the subsets.

3. **Calculating the total number of assignments:**
   Since each of the \(n\) elements in \(A\) can be assigned to the subsets in 7 different ways, the total number of ways to assign all elements is \(7^n\).

4. **Considering unordered triples:**
   The problem asks for unordered triples \((X, Y, Z)\). However, since the assignment of elements to subsets is inherently ordered, we do not need to further adjust for ordering.

Thus, the total number of unordered triples \((X, Y, Z)\) that satisfy \(X \cup Y \cup Z = A\) is \(7^n\).

The final answer is \(\boxed{7^n}\)