2017 AIME II Problems/Problem 1
Problem
Find the number of subsets of 
that are subsets of neither 
nor 
.
Solution 1
The number of subsets of a set with 
elements is 
. The total number of subsets of is equal to 
. The number of sets that are subsets of at least one of or can be found using complimentary counting. There are 
subsets of and subsets of . It is easy to make the mistake of assuming there are 
sets that are subsets of at least one of or , but the 
subsets of 
are overcounted. There are 
sets that are subsets of at least one of or , so there are 
subsets of that are subsets of neither nor . 
.
Solution 2
Upon inspection, a viable set must contain at least one element from both of the sets and . Since 4 and 5 are included in both of these sets, then they basically don't matter, i.e. if set A is a subset of both of those two then adding a 4 or a 5 won't change that fact. Thus, we can count the number of ways to choose at least one number from 1 to 3 and at least one number from 6 to 8, and then multiply that by the number of ways to add in 4 and 5. The number of subsets of a 3 element set is 
, but we want to exclude the empty set, giving us 7 ways to choose from 
or 
. We can take each of these 
sets and add in a 4 and/or a 5, which can be done in 4 different ways (by adding both, none, one, or the other one). Thus, the answer is 
.
Solution 3
The set of all subsets of 
that are disjoint with respect to 
and are not disjoint with respect to sets 
and 
will be named 
. There are 
members of set . The union of each member in and 
, 
, 
, and will be the members of set 
, which has 
members. 
Solution by User:a1b2
See Also
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Followed by Problem 2 | |
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