# Counting Sets

• Jul 25th 2007, 04:28 PM
tukeywilliams
Counting Sets
Determine with proof the number of ordered triples $(A_1, A_2, A_3)$ of sets which have the property (i) $A_1 \cup A_2 \cup A_3 = \{1,2,3,4,5,6,7,8,9,10 \}$ and (ii) $A_1 \cap A_2 \cap A_3 = \emptyset$. So basically this means that all the sets are disjoint.
• Jul 25th 2007, 05:55 PM
ThePerfectHacker
Quote:

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

Quote:

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

Think of a Venn diagram, see below.

Now the numbers $\{1,2,3,4,5,6,7,8,9,10\}$ can be placed anyway except into the shaded area. So we are placing $10$ "marbles" into 6 "boxes" (because there are 7 sections and we omit the middle section). The formula is:
${{6+10-1}\choose 10}$
• Jul 25th 2007, 06:05 PM
tukeywilliams
Could you also use matrices to solve this problem?
• Jul 25th 2007, 07:02 PM
ThePerfectHacker
Quote:

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.
• Jul 25th 2007, 07:07 PM
tukeywilliams
I was thinking that we could somehow use a binary relation. So there is some type of mapping (maybe a bijection?) between $A_1 \cup A_2 \cup A_3 = \{1, \ldots, 10\}$, $A_1 \cap A_2 \cap A_3 = \emptyset$ and some type of $10 \times 3$ matrix with $0,1$ so that there are no rows that are $\{000 \}$ or $\{111 \}$.
• Jul 25th 2007, 08:52 PM
tukeywilliams
I think there are $6$ possibilities for each row and therefore $6^{10}$ total possible triples?
• Jul 26th 2007, 03:34 AM
Plato
Quote:

Originally Posted by tukeywilliams
I think there are $6$ possibilities for each row and therefore $6^{10}$ total possible triples?

On a strict reading of this problem as stated, I agree that the answer is $6^{10}$.

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?