Difference between revisions of "2017 AIME II Problems/Problem 1"
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− | =Problem= | + | ==Problem== |
Find the number of 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>. | Find the number of 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>. | ||
− | =Solution= | + | ==Solution== |
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 complimentary 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 complimentary 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>. | ||
Revision as of 12:48, 23 March 2017
Problem
Find the number of subsets of that are subsets of neither
nor
.
Solution
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
.
.
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 0 |
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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