# Finding the number of triples given certain conditions.

Find the total number of ordered triples ($$A_1,A_2,A_3$$) such that $$A_1 \cup A_2 \cup A_3=\{1,2,3,4,5,6,7,8,9,10\}$$ and $$A_1\cap A_2\cap A_3=\emptyset.$$ $$\text{Attempt}.$$ From second condition, we know that there can't be any common element in any set. Also, the total number of elements should be $$10$$. Let's say we choose $$0$$ elements for set $$A_1$$,$$1$$ for $$A_2$$ and remaining for $$A_3$$. Next time we choose $$0$$ elements for $$A_1$$,$$2$$ for $$A_2$$ and remaining for $$A_3$$ . Using this logic I found number of sets as $$3!\left(\sum_{i=0}^{10}\left({10\choose i}\left(\sum_{j=0}^{10-i}{10-i\choose j}\right)\right)\right)$$. Where the term $$3!$$ is for total number of ways in which elements in $$3$$ groups can be permuted in $$3$$ distinct sets. Is this approach correct? Also how to find the required sum.

• math.stackexchange.com/questions/3255812/…
– cqfd
Jun 9, 2019 at 8:26
• @Thomas Shelby there is difference in the logic can you tell if my approach is correct if not where is it flawed? Jun 9, 2019 at 8:31

Your assumption that there can't be any common element in any set is incorrect. The fact that $$A_1 \cap A_2 \cap A_3 = \emptyset$$ means that no single element can be in all three sets. However, the sets \begin{align*} A_1 & = \{1, 2, 3, 4, 5\}\\ A_2 & = \{4, 6, 8, 9, 10\}\\ A_3 & = \{2, 3, 5, 7\} \end{align*} satisfy the requirement that $$A_1 \cap A_2 \cap A_3 = \emptyset$$ since no element is in all three sets. Notice that the condition $$A_1 \cup A_2 \cup A_3 = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$$ is also satisfied.

The requirements of the problem dictate that given $$n \in \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$$, $$n$$ satisfies one of the following six mutually exclusive conditions:

1. $$n \in A_1, n \in A_2, n \notin A_3$$
2. $$n \in A_1, n \notin A_2, n \in A_3$$
3. $$n \notin A_1, n \in A_2, n \in A_3$$
4. $$n \in A_1, n \notin A_1, n \notin A_3$$
5. $$n \notin A_1, n \in A_2, n \notin A_3$$
6. $$n \notin A_1, n \notin A_2, n \notin A_3$$

Hence, there are six possible ways of assigning each of the ten elements, or $$6^{10}$$ possible assignments.

Note: What you attempted to count is the number of ways of distributing the elements so that $$A_1 \cup A_2 \cup A_3 = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$$ is satisfied and $$A_1 \cap A_2 = A_1 \cap A_3 = A_2 \cap A_3 = \emptyset$$ Then each $$n \in \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$$ satisfies one of the following three conditions:

1. $$n \in A_1, n \notin A_1, n \notin A_3$$
2. $$n \notin A_1, n \in A_2, n \notin A_3$$
3. $$n \notin A_1, n \notin A_2, n \notin A_3$$

Under these conditions, there are $$3$$ possible ways to assign each of the $$10$$ elements, so there are $$3^{10}$$ possible ordered triples.