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Difference between revisions of "2017 AIME II Problems/Problem 1"

(Created page with "<math>\textbf{Problem 1}</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,...")
 
Line 4: Line 4:
 
<math>\textbf{Problem 1 Solution}</math>
 
<math>\textbf{Problem 1 Solution}</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>.
 
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>.
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=See Also=
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{{AIME box|year=2017|n=I|num=First Problem|num-a=2}}
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{{MAA Notice}}

Revision as of 12:46, 23 March 2017

$\textbf{Problem 1}$ Find the number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$.

$\textbf{Problem 1 Solution}$ The number of subsets of a set with $n$ elements is $2^n$. The total number of subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ is equal to $2^8$. The number of sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$ can be found using complimentary counting. There are $2^5$ subsets of $\{1, 2, 3, 4, 5\}$ and $2^5$ subsets of $\{4, 5, 6, 7, 8\}$. It is easy to make the mistake of assuming there are $2^5+2^5$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$, but the $2^2$ subsets of $\{4, 5\}$ are overcounted. There are $2^5+2^5-2^2$ sets that are subsets of at least one of $\{1, 2, 3, 4, 5\}$ or $\{4, 5, 6, 7, 8\}$, so there are $2^8-(2^5+2^5-2^2)$ subsets of $\{1, 2, 3, 4, 5, 6, 7, 8\}$ that are subsets of neither $\{1, 2, 3, 4, 5\}$ nor $\{4, 5, 6, 7, 8\}$. $2^8-(2^5+2^5-2^2)=\boxed{196}$.

See Also

2017 AIME I (Problems • Answer Key • Resources)
Preceded by
[[2017 AIME I Problems/Problem {{{num-b}}}|Problem {{{num-b}}}]] 		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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