Difference between revisions of "2017 AIME II Problems/Problem 1"
m (→Solution 1) |
m (→Solution 3) |
||
Line 10: | Line 10: | ||
==Solution 3 == | ==Solution 3 == | ||
− | 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> | + | This solution is very similar to Solution <math>2</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|a1b2]] | Solution by [[User:a1b2|a1b2]] |
Revision as of 21:09, 21 February 2018
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 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
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.