Answer: Ordered Triples of Sets Covering {1,...,10}

Key Idea / Intuition

Each element of $\{1, \ldots, 10\}$ independently "decides" which of the three sets it belongs to. The union condition says every element must appear in at least one set; the intersection condition says no element may appear in all three. So for each element, we need to count membership patterns that are neither "in none" nor "in all three" — a clean inclusion/exclusion on each element separately, and since elements are independent, just raise to the 10th power.

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Formal Proof / Solution

Step 1: Encode membership.

For each element $k \in \{1, \ldots, 10\}$, define a membership vector $(x_1, x_2, x_3) \in \{0,1\}^3$ where $x_i = 1$ iff $k \in A_i$.

There are $2^3 = 8$ possible vectors total.

Step 2: Apply the two conditions.

• $A_1 \cup A_2 \cup A_3 = \{1,\ldots,10\}$ means: for each $k$, at least one $x_i = 1$, i.e., $(0,0,0)$ is forbidden.
• $A_1 \cap A_2 \cap A_3 = \emptyset$ means: no $k$ lies in all three sets, i.e., $(1,1,1)$ is forbidden.

Step 3: Count valid patterns per element.

Each element independently has $8 - 2 = 6$ valid membership patterns (all patterns except $(0,0,0)$ and $(1,1,1)$).

Step 4: Total count.

Since the 10 elements choose independently:

$\text{Total} = 6^{10}$

Step 5: Express in required form.

$6^{10} = (2 \cdot 3)^{10} = 2^{10} \cdot 3^{10}$

So $a = 10,\ b = 10,\ c = 0,\ d = 0$, giving:

$\boxed{2^{10} \cdot 3^{10} \cdot 5^0 \cdot 7^0}$

Why this is beautiful: The problem looks like it might need complicated inclusion-exclusion over sets, but the key insight is that the two global conditions on the triple decouple completely into independent local conditions on each element. The answer $6^{10}$ falls out almost immediately once you see this.