Difference between revisions of "2017 AIME II Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
Upon inspection, a viable set must contain at least one element from both of the sets <math>\{1, 2, 3, 4, 5\}</math> and <math>\{4, 5, 6, 7, 8\}</math>. 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 <math>2^3=8</math>, but we want to exclude the empty set, giving us 7 ways to choose from <math>\{1, 2, 3\}</math> or <math>\{4, 5, 6\}</math>. We can take each of these <math>7 \times 7=49</math> 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 <math>49 \times 4=\boxed{196}</math>. | Upon inspection, a viable set must contain at least one element from both of the sets <math>\{1, 2, 3, 4, 5\}</math> and <math>\{4, 5, 6, 7, 8\}</math>. 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 <math>2^3=8</math>, but we want to exclude the empty set, giving us 7 ways to choose from <math>\{1, 2, 3\}</math> or <math>\{4, 5, 6\}</math>. We can take each of these <math>7 \times 7=49</math> 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 <math>49 \times 4=\boxed{196}</math>. | ||
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+ | ==Solution 3== | ||
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+ | Name a valid set <math>S</math> If its intersection with <math>\{4,5\}</math> is <math>\varnothing</math>. Its intersection with sets <math>\{1,2,3\}</math> and <math>\{6,7,8\}</math> must not be <math>\varnothing</math>. There are <math>7\cdot7=49</math> ways to choose these sets. The union of valid sets <math>S</math> and <math>\varnothing</math>, <math>\{4\}</math>, <math>\{5\}</math>, and <math>\{4,5\}</math> produce all the valid sets. There are <math>49\cdot4=\boxed{196}</math> sets in question. <math>\blacksquare</math> | ||
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+ | Solution by [[User: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 14:47, 29 March 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
Name a valid set If its intersection with is . Its intersection with sets and must not be . There are ways to choose these sets. The union of valid sets and , , , and produce all the valid sets. There are sets in question.
Solution by User: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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