1. ## Counting Sets

Determine with proof the number of ordered triples $\displaystyle (A_1, A_2, A_3)$ of sets which have the property (i) $\displaystyle A_1 \cup A_2 \cup A_3 = \{1,2,3,4,5,6,7,8,9,10 \}$ and (ii) $\displaystyle A_1 \cap A_2 \cap A_3 = \emptyset$. So basically this means that all the sets are disjoint.

2. So basically this means that all the sets are disjoint.
Not, necessarily. $\displaystyle A=\{1\},B=\{1,...,9\}, C=\{10\}$. But that actually makes the problem easier!

Originally Posted by tukeywilliams
Determine with proof the number of ordered triples $\displaystyle (A_1, A_2, A_3)$ of sets which have the property (i) $\displaystyle A_1 \cup A_2 \cup A_3 = \{1,2,3,4,5,6,7,8,9,10 \}$ and (ii) $\displaystyle A_1 \cap A_2 \cap A_3 = \emptyset$.
Think of a Venn diagram, see below.

Now the numbers $\displaystyle \{1,2,3,4,5,6,7,8,9,10\}$ can be placed anyway except into the shaded area. So we are placing $\displaystyle 10$ "marbles" into 6 "boxes" (because there are 7 sections and we omit the middle section). The formula is:
$\displaystyle {{6+10-1}\choose 10}$

3. Could you also use matrices to solve this problem?

4. Originally Posted by tukeywilliams
Could you also use matrices to solve this problem?
I really have no idea? I am not a Combinatorics expert .

Perhaps, you can phrase this problem as a Graph Theory problem and then arrive at your answer by computing the adjancency matrix. But I have no idea how to do that.

5. I was thinking that we could somehow use a binary relation. So there is some type of mapping (maybe a bijection?) between $\displaystyle A_1 \cup A_2 \cup A_3 = \{1, \ldots, 10\}$,$\displaystyle A_1 \cap A_2 \cap A_3 = \emptyset$ and some type of $\displaystyle 10 \times 3$ matrix with $\displaystyle 0,1$ so that there are no rows that are $\displaystyle \{000 \}$ or $\displaystyle \{111 \}$.

6. I think there are $\displaystyle 6$ possibilities for each row and therefore $\displaystyle 6^{10}$ total possible triples?

7. Originally Posted by tukeywilliams
I think there are $\displaystyle 6$ possibilities for each row and therefore $\displaystyle 6^{10}$ total possible triples?
On a strict reading of this problem as stated, I agree that the answer is $\displaystyle 6^{10}$.

However, are you sure that you have stated this question correctly?
As stated, the question has several different readings.
I have seen this sort of question many times: “How many ways can {0,1,2,…,8,9} be partitioned into three non-empty, pair-wise disjoint sets?”
Could that be what was meant by this question?