1
$\begingroup$

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.

$\endgroup$
2
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.