 Let Ω = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. What is the number of ordered triples (A1, A2, A3) of sets
which have the property that
(i) A1 ∪ A2 ∪ A3 = Ω, and
(ii) A1 ∩ A2 ∩ A3 = Ø

We are tasked with determining the number of ordered triples \((A_1, A_2, A_3)\) of sets, such that:

1. \(A_1 \cup A_2 \cup A_3 = \Omega = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\),
2. \(A_1 \cap A_2 \cap A_3 = \emptyset\) (i.e., no element of \(\Omega\) is in all three sets simultaneously).

### Step-by-step solution:

Each element of \(\Omega\) can be placed in one of several possible configurations of the sets \(A_1, A_2, A_3\) such that both conditions are satisfied:

1. **Condition 1: \(A_1 \cup A_2 \cup A_3 = \Omega\)**:  
   Every element must belong to at least one of the sets \(A_1, A_2,\) or \(A_3\). Thus, each element of \(\Omega\) must be assigned to one or more of the sets.

2. **Condition 2: \(A_1 \cap A_2 \cap A_3 = \emptyset\)**:  
   No element can belong to all three sets simultaneously. This condition restricts the configurations where an element would be in \(A_1 \cap A_2 \cap A_3\).

### Analyzing the choices for each element:

For each element of \(\Omega\), we have to decide in which subsets of \(A_1, A_2, A_3\) the element will belong. The possibilities are as follows:

- The element can be in \(A_1\) only.
- The element can be in \(A_2\) only.
- The element can be in \(A_3\) only.
- The element can be in both \(A_1\) and \(A_2\).
- The element can be in both \(A_2\) and \(A_3\).
- The element can be in both \(A_1\) and \(A_3\).
- The element can be in \(A_1\) and \(A_2\) and also in \(A_3\) **but not all three simultaneously** (this case is prohibited).

Thus, the number of valid configurations for each element is 7 (since we cannot have the element in all three sets simultaneously).

### Total number of configurations:

Since there are 10 elements in \(\Omega\), and each element has 7 valid choices, the total number of ordered triples \((A_1, A_2, A_3)\) is:

\[
7^{10}
\]

This is the final answer.

Would you like more details or further explanations? Here are 5 related questions that may interest you:

1. How does this approach relate to the principle of inclusion-exclusion?
2. Can this method be extended to more than three sets?
3. How would the result change if the condition \(A_1 \cup A_2 \cup A_3 = \Omega\) were relaxed?
4. How does this problem relate to the concept of set partitions?
5. What are other combinatorial counting methods that could be applied to this type of problem?

**Tip:** Always be mindful of conditions that restrict the overlap between sets, as they can significantly reduce the number of possible configurations.

Let Ω = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. What is the number of ordered triples (A1, A2, A3) of sets which have the property that (i) A1 ∪ A2 ∪ A3 = Ω, and (ii) A1 ∩ A2 ∩ A3 = Ø?

Learn how to calculate the number of ordered triples of sets (A1, A2, A3) that satisfy the conditions A1 ∪ A2 ∪ A3 = Ω and A1 ∩ A2 ∩ A3 = Ø. This problem covers combinatorial counting techniques and set theory concepts. Suitable for students in Grades 10-12.

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