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.
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$\begingroup$ math.stackexchange.com/questions/3255812/… $\endgroup$– cqfdJun 9, 2019 at 8:26
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$\begingroup$ @Thomas Shelby there is difference in the logic can you tell if my approach is correct if not where is it flawed? $\endgroup$– Archis WelankarJun 9, 2019 at 8:31
1 Answer
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:
- $n \in A_1, n \in A_2, n \notin A_3$
- $n \in A_1, n \notin A_2, n \in A_3$
- $n \notin A_1, n \in A_2, n \in A_3$
- $n \in A_1, n \notin A_1, n \notin A_3$
- $n \notin A_1, n \in A_2, n \notin A_3$
- $n \notin A_1, n \notin A_2, n \notin A_3$
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:
- $n \in A_1, n \notin A_1, n \notin A_3$
- $n \notin A_1, n \in A_2, n \notin A_3$
- $n \notin A_1, n \notin A_2, n \notin A_3$
Under these conditions, there are $3$ possible ways to assign each of the $10$ elements, so there are $3^{10}$ possible ordered triples.
