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Difference between revisions of "2017 AIME II Problems/Problem 1"
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=Problem=
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==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>.
  
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=Solution=
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==Solution 1==
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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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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 complementary 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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==Solution 2==
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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>\{6, 7, 8\}</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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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>
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Solution by [[User:a1b2|a1b2]]
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==Solution 4==
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Consider that we are trying to figure out how many subsets are possible of <math>\{1,2,3,4,5,6,7,8\}</math> that are not in violation of the two subsets <math>\{1,2,3,4,5\}</math> and <math>\{4,5,6,7,8\}</math>. Assume that the number of numbers we pick from the subset <math>\{1,2,3,4,5,6,7,8\}</math> is <math>n</math>. Thus, we can compute this problem with simple combinatorics:
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If <math>n=1</math>, <math>\binom{8}{1}</math> <math>-</math> (<math>\binom{5}{1}</math> + <math>\binom{5}{1}</math> <math>-  2</math>) [subtract <math>2</math> to eliminate the overcounting of the subset <math>\{4\}</math> or <math>\{5\}</math>] = <math>8 - 8</math> = <math>0</math>
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If <math>n=2</math>, <math>\binom{8}{2}</math> <math>-</math> (<math>\binom{5}{2}</math> + <math>\binom{5}{2}</math> <math>-  1</math>) [subtract <math>1</math> to eliminate the overcounting of the subset <math>\{4,5\}</math>] = <math>28 - 19</math> = <math>9</math>
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If <math>n=3</math>, <math>\binom{8}{3}</math> <math>-</math> (<math>\binom{5}{3}</math> + <math>\binom{5}{3}</math>) = <math>56 - 20</math> = <math>36</math>
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If <math>n=4</math>, <math>\binom{8}{4}</math> <math>-</math> (<math>\binom{5}{4}</math> + <math>\binom{5}{4}</math>) = <math>70 - 10</math> = <math>60</math>
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If <math>n=5</math>, <math>\binom{8}{5}</math> <math>-</math> (<math>\binom{5}{5}</math> + <math>\binom{5}{5}</math>) = <math>56 - 2</math> = <math>54</math>
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If <math>n>5</math>, then the set <math>\{1,2,3,4,5,6,7,8\}</math> is never in violation of the two subsets <math>\{1,2,3,4,5\}</math> and <math>\{4,5,6,7,8\}</math>. Thus,
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If <math>n=6</math>, <math>\binom{8}{6}</math> = <math>28</math>
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If <math>n=7</math>, <math>\binom{8}{7}</math> = <math>8</math>
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If <math>n=8</math>, <math>\binom{8}{8}</math> = <math>1</math>
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Adding these together, our solution becomes <math>0</math> + <math>9</math> + <math>36</math> + <math>60</math> + <math>54</math> + <math>28</math> + <math>8</math> + <math>1</math> <math>=\boxed{196}</math>
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Solution by IronicNinja~
  
 
=See Also=
 
=See Also=
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{{AIME box|year=2017|n=I|num-b=0|num-a=2}}
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{{AIME box|year=2017|n=II|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:02, 29 January 2024

Problem

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 $\{4, 5, 6, 7, 8\}$.

Solution 1

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 complementary 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 $\{4, 5, 6, 7, 8\}$. $2^8-(2^5+2^5-2^2)=\boxed{196}$.

Solution 2

Upon inspection, a viable set must contain at least one element from both of the sets $\{1, 2, 3, 4, 5\}$ and $\{4, 5, 6, 7, 8\}$. 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 $2^3=8$, but we want to exclude the empty set, giving us 7 ways to choose from $\{1, 2, 3\}$ or $\{6, 7, 8\}$. We can take each of these $7 \times 7=49$ 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 $49 \times 4=\boxed{196}$.

Solution 3

This solution is very similar to Solution $2$. The set of all subsets of $\{1,2,3,4,5,6,7,8\}$ that are disjoint with respect to $\{4,5\}$ and are not disjoint with respect to the complements of sets (and therefore not a subset of) $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$ will be named $S$, which has $7\cdot7=49$ members. The union of each member in $S$ and the $2^2=4$ subsets of $\{4,5\}$ will be the members of set $Z$, which has $49\cdot4=\boxed{196}$ members. $\blacksquare$

Solution by a1b2

Solution 4

Consider that we are trying to figure out how many subsets are possible of $\{1,2,3,4,5,6,7,8\}$ that are not in violation of the two subsets $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$. Assume that the number of numbers we pick from the subset $\{1,2,3,4,5,6,7,8\}$ is $n$. Thus, we can compute this problem with simple combinatorics:

If $n=1$, $\binom{8}{1}$ $-$ ($\binom{5}{1}$ + $\binom{5}{1}$ $- 2$) [subtract $2$ to eliminate the overcounting of the subset $\{4\}$ or $\{5\}$] = $8 - 8$ = $0$

If $n=2$, $\binom{8}{2}$ $-$ ($\binom{5}{2}$ + $\binom{5}{2}$ $- 1$) [subtract $1$ to eliminate the overcounting of the subset $\{4,5\}$] = $28 - 19$ = $9$

If $n=3$, $\binom{8}{3}$ $-$ ($\binom{5}{3}$ + $\binom{5}{3}$) = $56 - 20$ = $36$

If $n=4$, $\binom{8}{4}$ $-$ ($\binom{5}{4}$ + $\binom{5}{4}$) = $70 - 10$ = $60$

If $n=5$, $\binom{8}{5}$ $-$ ($\binom{5}{5}$ + $\binom{5}{5}$) = $56 - 2$ = $54$

If $n>5$, then the set $\{1,2,3,4,5,6,7,8\}$ is never in violation of the two subsets $\{1,2,3,4,5\}$ and $\{4,5,6,7,8\}$. Thus,

If $n=6$, $\binom{8}{6}$ = $28$

If $n=7$, $\binom{8}{7}$ = $8$

If $n=8$, $\binom{8}{8}$ = $1$

Adding these together, our solution becomes $0$ + $9$ + $36$ + $60$ + $54$ + $28$ + $8$ + $1$ $=\boxed{196}$

Solution by IronicNinja~

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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