The question:
Determine the amount of ordered triples of sets $(A_1,A_2,A_3)$ such that:
- $A_1 \cup A_2 \cup A_3 =\{1,2,3....,10\}$,
- $A_i \cap A_j=\emptyset$ for $i\neq j$.
My approach:
Consider the following variables: $p_1,....,p_{10}$.
I want to divide them over $A_1,...,A_3$ such that the intersection of the three sets is empty. Since the order of the $p_i$ does not matter in this state, we can divide them by order so we start with set $A_1$ put numbers in there (or not), for instance $p_1,p_2$ then we switch to $A_2$ for instance $p_3,p_4,p_5$ and then we put the rest in $A_3$.
We can see this type of division of the $p_i$ over the $A_{i}$ as a chain of 12 events where 2 of them are the event 'switch', so ${12 \choose 2}$ possibilities.
Now we have to assign to $p_i$ to a number. We can say that $p_{1}$ has 10 options, $p_{2}$ 9 etc. That brings us to ${12 \choose 2} \times 10!$
Lastly we have to see how many options there are to order $A_{1},..,A_{3}$. This is $3!$.
So the final answer would be:
$${12 \choose 2} \times 10! \times 3!$$
Is my approach correct? Thanks in advance This question is not a duplicate since the other question doesn't mention an approach/check my proof.