Difference between revisions of "2017 AIME II Problems/Problem 1"
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The set of all subsets of <math>\{1,2,3,4,5,6,7,8\}</math> that are disjoint with respect to <math>\{4,5\}</math> and are not disjoint with respect to the complements of sets (and therefore not a subset of) <math>\{1,2,3,4,5\}</math> and <math>\{4,5,6,7,8\}</math> will be named <math>S</math>, which has <math>7\cdot7=49</math> members. The union of each member in <math>S</math> and the <math>2^2=4</math> subsets of <math>\{4,5\}</math> will be the members of set <math>Z</math>, which has <math>49\cdot4=\boxed{196}</math> members. <math>\blacksquare</math> | The set of all subsets of <math>\{1,2,3,4,5,6,7,8\}</math> that are disjoint with respect to <math>\{4,5\}</math> and are not disjoint with respect to the complements of sets (and therefore not a subset of) <math>\{1,2,3,4,5\}</math> and <math>\{4,5,6,7,8\}</math> will be named <math>S</math>, which has <math>7\cdot7=49</math> members. The union of each member in <math>S</math> and the <math>2^2=4</math> subsets of <math>\{4,5\}</math> will be the members of set <math>Z</math>, which has <math>49\cdot4=\boxed{196}</math> members. <math>\blacksquare</math> | ||
− | Solution by [[User:a1b2]] | + | Solution by [[User:a1b2|a1b2]] |
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2017|n=II|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:13, 1 April 2017
Contents
[hide]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 (Gets straight to the point)
The set of all subsets of that are disjoint with respect to
and are not disjoint with respect to the complements of sets (and therefore not a subset of)
and
will be named
, which has
members. The union of each member in
and the
subsets of
will be the members of set
, which has
members.
Solution by 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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