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
Name a valid set If its intersection with
is
. Its intersection with sets
and
must not be
. There are
ways to choose these sets. The union of valid sets
and
,
,
, and
produce all the valid sets. There are
sets in question.
Solution by User:a1b2
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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