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
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
