Putnam 1985, A1

Determine, with proof, the number of ordered triples ( A 1 , A 2 , A 3 ) (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 , 10 } A_{1} ∪ A_{2} ∪ A_{3} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} , and
(ii) A 1 A 2 A 3 = A_{1} ∩ A_{2} ∩ A_{3} = ∅ ,

where denotes the empty set.


The answer is 60466176.

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3 solutions

Pranshu Gaba
Jul 17, 2015

Each element 1 , 2 , 3 , 10 1, 2, 3, \ldots 10 must belong to exactly one of the 6 6 regions. By Rule of Product , it can be done in 6 10 = 60466176 6^{10} = \boxed{60466176} ways. _\square

Kenny Lau
Jul 19, 2015

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 A_1 . There are three sets, giving 2 × 2 × 2 = 8 2\times2\times2=8 choices. However, two aforementioned choices are discarded, leaving 6 6 choices only for one number.

The same is true for all the integers from 1 to 10.

Therefore, the answer is 6 6 multiplied itself 10 10 times, or 6 10 = 60466176 6^{10}=\mbox{60466176} times.

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