Difference between revisions of "2017 AIME II Problems/Problem 1"
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==Solution 1== | ==Solution 1== | ||
The number of subsets of a set with <math>n</math> elements is <math>2^n</math>. The total number of subsets of <math>\{1, 2, 3, 4, 5, 6, 7, 8\}</math> is equal to <math>2^8</math>. The number of sets that are subsets of at least one of <math>\{1, 2, 3, 4, 5\}</math> or <math>\{4, 5, 6, 7, 8\}</math> can be found using complementary counting. There are <math>2^5</math> subsets of <math>\{1, 2, 3, 4, 5\}</math> and <math>2^5</math> subsets of <math>\{4, 5, 6, 7, 8\}</math>. It is easy to make the mistake of assuming there are <math>2^5+2^5</math> sets that are subsets of at least one of <math>\{1, 2, 3, 4, 5\}</math> or <math>\{4, 5, 6, 7, 8\}</math>, but the <math>2^2</math> subsets of <math>\{4, 5\}</math> are overcounted. There are <math>2^5+2^5-2^2</math> sets that are subsets of at least one of <math>\{1, 2, 3, 4, 5\}</math> or <math>\{4, 5, 6, 7, 8\}</math>, so there are <math>2^8-(2^5+2^5-2^2)</math> subsets of <math>\{1, 2, 3, 4, 5, 6, 7, 8\}</math> that are subsets of neither <math>\{1, 2, 3, 4, 5\}</math> nor <math>\{4, 5, 6, 7, 8\}</math>. <math>2^8-(2^5+2^5-2^2)=\boxed{196}</math>. | The number of subsets of a set with <math>n</math> elements is <math>2^n</math>. The total number of subsets of <math>\{1, 2, 3, 4, 5, 6, 7, 8\}</math> is equal to <math>2^8</math>. The number of sets that are subsets of at least one of <math>\{1, 2, 3, 4, 5\}</math> or <math>\{4, 5, 6, 7, 8\}</math> can be found using complementary counting. There are <math>2^5</math> subsets of <math>\{1, 2, 3, 4, 5\}</math> and <math>2^5</math> subsets of <math>\{4, 5, 6, 7, 8\}</math>. It is easy to make the mistake of assuming there are <math>2^5+2^5</math> sets that are subsets of at least one of <math>\{1, 2, 3, 4, 5\}</math> or <math>\{4, 5, 6, 7, 8\}</math>, but the <math>2^2</math> subsets of <math>\{4, 5\}</math> are overcounted. There are <math>2^5+2^5-2^2</math> sets that are subsets of at least one of <math>\{1, 2, 3, 4, 5\}</math> or <math>\{4, 5, 6, 7, 8\}</math>, so there are <math>2^8-(2^5+2^5-2^2)</math> subsets of <math>\{1, 2, 3, 4, 5, 6, 7, 8\}</math> that are subsets of neither <math>\{1, 2, 3, 4, 5\}</math> nor <math>\{4, 5, 6, 7, 8\}</math>. <math>2^8-(2^5+2^5-2^2)=\boxed{196}</math>. | ||
+ | |||
+ | ==Solution 1.1 ( PIE Simplified )== | ||
+ | Note that by Principle of Inclusion and Exclusion, the total number of subsets must be <math>2^8-2^5-2^5+2^2</math> as denoted by above. Thus our answer is <math>64(3)+4 = \boxed{196}</math> | ||
==Solution 2== | ==Solution 2== |
Revision as of 12:13, 30 April 2021
Contents
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 complementary 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 1.1 ( PIE Simplified )
Note that by Principle of Inclusion and Exclusion, the total number of subsets must be as denoted by above. Thus our answer is
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
This solution is very similar to Solution . 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
Solution 4
Consider that we are trying to figure out how many subsets are possible of that are not in violation of the two subsets
and
. Assume that the number of numbers we pick from the subset
is
. Thus, we can compute this problem with simple combinatorics:
If ,
(
+
) [subtract
to eliminate the overcounting of the subset
or
] =
=
If ,
(
+
) [subtract
to eliminate the overcounting of the subset
] =
=
If ,
(
+
) =
=
If ,
(
+
) =
=
If ,
(
+
) =
=
If , then the set
is never in violation of the two subsets
and
. Thus,
If ,
=
If ,
=
If ,
=
Adding these together, our solution becomes +
+
+
+
+
+
+
Solution by IronicNinja~
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.