Determine, with proof, the number of ordered triples ( A 1 , A 2 , A 3 ) of sets which have the property that
(i)
A
1
∪
A
2
∪
A
3
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
,
9
,
1
0
}
, and
(ii)
A
1
∩
A
2
∩
A
3
=
∅
,
where ∅ denotes the empty set.
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This problem is done on the perspective of the numbers.
This problem is done by determining which set the numbers can be in. Let's take 1 as an example. It cannot be in neither sets, or (i) will not hold; it cannot be in all sets, or (ii) will not hold.
It either belongs or does not belong in set A 1 . There are three sets, giving 2 × 2 × 2 = 8 choices. However, two aforementioned choices are discarded, leaving 6 choices only for one number.
The same is true for all the integers from 1 to 10.
Therefore, the answer is 6 multiplied itself 1 0 times, or 6 1 0 = 60466176 times.
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Each element 1 , 2 , 3 , … 1 0 must belong to exactly one of the 6 regions. By Rule of Product , it can be done in 6 1 0 = 6 0 4 6 6 1 7 6 ways. □