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Difference between revisions of "2018 AIME I Problems/Problem 9"

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How many distinct subsets of <math>S={1, 2, ..., 19, 20}</math> are there such that there are two distinct elements that sum to <math>16</math> and two distinct elements that sum to <math>24</math>? Two examples are <math>{6, 10, 20, 18}</math> and <math>{3, 5, 13, 19}</math>.
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==Problem==
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Find the number of four-element subsets of <math>\{1,2,3,4,\dots, 20\}</math> with the property that two distinct elements of a subset have a sum of <math>16</math>, and two distinct elements of a subset have a sum of <math>24</math>. For example, <math>\{3,5,13,19\}</math> and <math>\{6,10,20,18\}</math> are two such subsets.
  
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Feel free to correct this problem if the wording isn't verbatim (I know it isn't).
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==Solution 1==
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This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set <math>\{a, b, c, d\}</math>.
  
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==Solutions==
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Note that there are only two cases: 1 where <math>a + b = 16</math> and <math>c + d = 24</math> or 2 where <math>a + b = 16</math> and <math>a + c = 24</math>. Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you <math>a=d</math>, which cannot be true.
  
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==Solution Exclusion Awareness==
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Case 1.
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This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set <math>{a, b, c, d}</math>.
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This is probably the simplest: just make a list of possible combinations for <math>\{a, b\}</math> and <math>\{c, d\}</math>. We get <math>\{1, 15\}\dots\{7, 9\}</math> for the first and <math>\{4, 20\}\dots\{11, 13\}</math> for the second. That appears to give us <math>7*8=56</math> solutions, right? NO. Because elements can't repeat, take out the supposed sets
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<cmath>\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 12, 4, 20\}, \{5, 11, 5, 19\},</cmath><cmath>\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}</cmath> That's ten cases gone. So <math>46</math> for Case 1.
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Case 2.
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We can look for solutions by listing possible <math>a</math> values and filling in the blanks. Start with <math>a=4</math>, as that is the minimum. We find <math>\{4, 12, 20, ?\}</math>, and likewise up to <math>a=15</math>. But we can't have <math>a=8</math> or <math>a=12</math> because <math>a=b</math> or <math>a=c</math>, respectively! Now, it would seem like there are <math>10</math> values for <math>a</math> and <math>17</math> unique values for each <math>?</math>, giving a total of <math>170</math>, but that is once again not true because there are some repeated values!
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There are two cases of overcounting:
 +
 
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case 1) (5,11,13,19) & (5.11.19.13)
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The same is for (6,10,14,18) and (7,9,15,17)
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case 2) those that have the same b and c values
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this case includes:
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(1,15,9,7) and (7,9,15,1)
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(2,14,10,6) and (6,10,14,2)
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(3,13,11,5) and (5,11,13,3)
  
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Note that there are only two cases: 1 where <math>a + b = 16</math> and <math>c + d = 24</math> or 2 where <math>a + b = 16</math> and <math>a + c = 24</math>. Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you <math>a=d</math>, which is absurd.
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So we need to subtract 6 overcounts.
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So, that's <math>164</math> for Case 2.
  
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Case 1.
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Total gives <math>\boxed{210}</math>.
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This is probably the simplest: just make a list of possible combinations for <math>{a, b}</math> and <math>{c, d}</math>. We get <math>{1, 15}...{7, 9}</math> for the first and <math>{4, 20}...{11, 13}</math> for the second. That appears to give us <math>7*8=56</math> solutions, right? NO. Because elements can't repeat, take out the supposed sets <math>{a, b, c, d}</math> of <math>{1, 15, 9, 15}, {2, 14, 10, 14}, {3, 13, 11, 13}, {4, 16, 4, 20}, {5, 11, 5, 19}, {5, 11, 11, 13}, {6, 10, 6, 18}, {6, 10, 10, 14}, {7, 9, 9, 15}, {7, 9, 7, 17}</math>. That's ten cases gone. So <math>46</math> for Case 1.
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-expiLnCalc
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==Solution 2==
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Let's say our four elements in our subset are <math>a,b,c,d</math>. We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.
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<math>\textrm{Case } 1 \textrm{:}</math> <math>a+b = 16</math> and <math>c+d = 24</math>.  
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List out possibilities for <math>a+b</math> <math>(\text{i.e. } 1+15, 2+14, 3+13 \text{ etc.})</math> but don't list <math>8+8</math> because those are the same elements and that is restricted.  
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Then list out the possibilities for <math>c+d \text{ }(\text{i.e. } 4+20, 5+19, 6+18, \text{ etc.})</math> but don't list <math>12+12</math> because they are the same elements.
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This will give you <math>7 \cdot 8</math> elements, which is <math>56</math>. However, as stated above, we have overlap. Just count starting from <math>a+b</math>.  <math>15,14,13,11,10,9,7,6,5,4</math> all overlap once, which is <math>10</math>, thus <math>56 - 10 = 46</math> cases in this case. Note that <math>12</math> wasn't included because again, if <math>c+d = 24</math>, <math>c</math> and <math>d</math> cannot be <math>12</math>.
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<math>\textrm{Case } 2 \textrm{:}</math> <math>a+b = 16</math> and <math>b+c = 24</math>.
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Here, <math>b</math> is included in both equations. We can easily see that <math>a, b, c</math> will never equal each other.
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Furthermore, there are 17 choices for <math>d</math> (<math>20 - 3</math> included elements) for each <math>b</math>. Listing out the possible <math>b</math>s, we go from <math>15,14,13,11,10,9,7,6,5,4</math>. Do not include <math>8</math> or <math>12</math> because if they are included, then <math>a/c</math> will be the same as <math>b</math>, which is restricted.  
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There are <math>10</math> options there, and thus <math>10 \cdot 17 = 170</math>. But, note that if <math>d = b+8</math>, <math>a+d = a+b+8 = 24</math>, and so we have a double-counted set. Starting with <math>b=15</math>, we have <math>15, 14, 13, 11, 10, 9</math> (where <math>d</math> is <math>7, 6, 5, 3, 2, 1)</math>. That means there are <math>6</math> double-counted cases. Thus <math>170 - 6 = 164</math> cases in this case.
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Adding these up, we get <math>46+164 = \boxed{210}.</math>
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~IronicNinja
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~<math>\LaTeX</math> by AlcBoy1729
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~Formatted by ojaswupadhyay and phoenixfire
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==Solution 3 (Official MAA)==
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There are two types of <math>\{a,b,c,d\} \subseteq \{1,2,3,4\dots,20\}</math> that have the needed property. There is either an assignment of distinct values for <math>a,\,b,\,c,</math> and <math>d</math> such that <math>a+b=16</math> and <math>c+d=24</math> or an assignment such that <math>a+b=16</math> and <math>a+c=24.</math> These two types are mutually exclusive because <math>c+d=24</math> and <math>a+c=24</math> imply that <math>a=d.</math> For the first type, there are <math>7</math> choices for <math>\{a,b\},</math> namely <math>\{1,15\},\,\{2,14\},\,\{3,13\},\,\{4,12\},\,\{5,11\},\,\{6,10\},</math> and <math>\{7,9\},</math> and there are <math>8</math> choices for <math>\{c,d\},</math> namely <math>\{4,20\},\,\{5,19\},\,\{6,18\},\,\{7,17\},\,\{8,16\},\,\{9,15\},\{10,14\},</math> and <math>\{11,13\}.</math> Thus a four-element subset of the first type can be formed by taking the union of one of <math>7</math> two-element subsets with one of <math>8</math> two-element subsets as long as these two subsets are disjoint. There are <math>10</math> such pairings that are not disjoint out of the <math>7\cdot 8=56</math> pairings, so there are <math>56-10=46</math> subsets of the first type.
  
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Case 2.
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For subsets of the second type, there are <math>10</math> choices for the value of <math>a</math> <math>(4,\,5,\,6,\,7,\,9,\,10,\,11,\,13,\,14,\,15)</math> such that <math>b=16-a</math> and <math>c=24-a</math> can be two other elements of the subset. Note that in each of these cases, <math>c-b=(24-a)-(16-a)=8.</math> For each of these, there are <math>20-3=17</math> other values that can be chosen for the element <math>d</math> in the subset. But <math>10\cdot 17=170</math> counts some subsets more than once. In particular, a subset is counted twice if <math>b+d=24</math> or <math>c+d=16</math>. In such cases either <math>d=a+8</math> or <math>d=a-8</math>. There are exactly <math>6</math> subsets where the role of <math>a</math> can be played by two different elements of the set. They are <math>\{1,7,9,15\},\,\{2,6,10,14\},\,\{3,5,11,13\},\,\{5,11,13,19\},\,\{6,10,14,18\},</math> and <math>\{7,9,15,17\}</math>. Thus there are <math>170-6=164</math> subsets of the second type.
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We can look for solutions by listing possible <math>a</math> values and filling in the blanks. Start with <math>a=4</math>, as that is the minimum. We find <math>{4, 12, 20, ?}</math>, and likewise up to <math>a=15</math>. But we can't have <math>a=8</math> or <math>a=12</math> because <math>a=b</math> or <math>a=c</math>, respectively! Now, it would seem like there are <math>10</math> values for <math>a</math> and <math>17</math> unique values for each <math>?</math>, giving a total of <math>170</math>, but that is once again not true because there are some repeated values! We can subtract 1 from all pairs of sets that have two elements in common, because those can give us identical sets. There are 3 pairs about <math>a=8</math> and 3 pairs about <math>a=12</math>, meaning we lose <math>6</math>. That's <math>164</math> for Case 2.
 
  
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Total gives <math>\boxed{210}</math>. Lesson for this problem: Never be scared to attempt an AIME problem. You will oftentimes get it in ~10 minutes.
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In all, there are <math>46+164=210</math> subsets with the required property.
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{{AIME box|year=2018|n=I|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 22:32, 5 February 2022

Problem

Find the number of four-element subsets of $\{1,2,3,4,\dots, 20\}$ with the property that two distinct elements of a subset have a sum of $16$, and two distinct elements of a subset have a sum of $24$. For example, $\{3,5,13,19\}$ and $\{6,10,20,18\}$ are two such subsets.

Solution 1

This problem is tricky because it is the capital of a few "bashy" calculations. Nevertheless, the process is straightforward. Call the set $\{a, b, c, d\}$.

Note that there are only two cases: 1 where $a + b = 16$ and $c + d = 24$ or 2 where $a + b = 16$ and $a + c = 24$. Also note that there is no overlap between the two situations! This is because if they overlapped, adding the two equations of both cases and canceling out gives you $a=d$, which cannot be true.

Case 1. This is probably the simplest: just make a list of possible combinations for $\{a, b\}$ and $\{c, d\}$. We get $\{1, 15\}\dots\{7, 9\}$ for the first and $\{4, 20\}\dots\{11, 13\}$ for the second. That appears to give us $7*8=56$ solutions, right? NO. Because elements can't repeat, take out the supposed sets \[\{1, 15, 9, 15\}, \{2, 14, 10, 14\}, \{3, 13, 11, 13\}, \{4, 12, 4, 20\}, \{5, 11, 5, 19\},\]\[\{5, 11, 11, 13\}, \{6, 10, 6, 18\}, \{6, 10, 10, 14\}, \{7, 9, 9, 15\}, \{7, 9, 7, 17\}\] That's ten cases gone. So $46$ for Case 1.

Case 2. We can look for solutions by listing possible $a$ values and filling in the blanks. Start with $a=4$, as that is the minimum. We find $\{4, 12, 20, ?\}$, and likewise up to $a=15$. But we can't have $a=8$ or $a=12$ because $a=b$ or $a=c$, respectively! Now, it would seem like there are $10$ values for $a$ and $17$ unique values for each $?$, giving a total of $170$, but that is once again not true because there are some repeated values! There are two cases of overcounting:

case 1) (5,11,13,19) & (5.11.19.13)

The same is for (6,10,14,18) and (7,9,15,17)

case 2) those that have the same b and c values

this case includes:

(1,15,9,7) and (7,9,15,1)

(2,14,10,6) and (6,10,14,2)

(3,13,11,5) and (5,11,13,3)

So we need to subtract 6 overcounts. So, that's $164$ for Case 2.

Total gives $\boxed{210}$.

-expiLnCalc

Solution 2

Let's say our four elements in our subset are $a,b,c,d$. We have two cases. Note that the order of the elements / the element letters themselves don't matter since they are all on equal grounds at the start.


$\textrm{Case } 1 \textrm{:}$ $a+b = 16$ and $c+d = 24$.

List out possibilities for $a+b$ $(\text{i.e. } 1+15, 2+14, 3+13 \text{ etc.})$ but don't list $8+8$ because those are the same elements and that is restricted.

Then list out the possibilities for $c+d \text{ }(\text{i.e. } 4+20, 5+19, 6+18, \text{ etc.})$ but don't list $12+12$ because they are the same elements.

This will give you $7 \cdot 8$ elements, which is $56$. However, as stated above, we have overlap. Just count starting from $a+b$. $15,14,13,11,10,9,7,6,5,4$ all overlap once, which is $10$, thus $56 - 10 = 46$ cases in this case. Note that $12$ wasn't included because again, if $c+d = 24$, $c$ and $d$ cannot be $12$.


$\textrm{Case } 2 \textrm{:}$ $a+b = 16$ and $b+c = 24$.

Here, $b$ is included in both equations. We can easily see that $a, b, c$ will never equal each other.

Furthermore, there are 17 choices for $d$ ($20 - 3$ included elements) for each $b$. Listing out the possible $b$s, we go from $15,14,13,11,10,9,7,6,5,4$. Do not include $8$ or $12$ because if they are included, then $a/c$ will be the same as $b$, which is restricted.

There are $10$ options there, and thus $10 \cdot 17 = 170$. But, note that if $d = b+8$, $a+d = a+b+8 = 24$, and so we have a double-counted set. Starting with $b=15$, we have $15, 14, 13, 11, 10, 9$ (where $d$ is $7, 6, 5, 3, 2, 1)$. That means there are $6$ double-counted cases. Thus $170 - 6 = 164$ cases in this case.

Adding these up, we get $46+164 = \boxed{210}.$

~IronicNinja ~$\LaTeX$ by AlcBoy1729 ~Formatted by ojaswupadhyay and phoenixfire

Solution 3 (Official MAA)

There are two types of $\{a,b,c,d\} \subseteq \{1,2,3,4\dots,20\}$ that have the needed property. There is either an assignment of distinct values for $a,\,b,\,c,$ and $d$ such that $a+b=16$ and $c+d=24$ or an assignment such that $a+b=16$ and $a+c=24.$ These two types are mutually exclusive because $c+d=24$ and $a+c=24$ imply that $a=d.$ For the first type, there are $7$ choices for $\{a,b\},$ namely $\{1,15\},\,\{2,14\},\,\{3,13\},\,\{4,12\},\,\{5,11\},\,\{6,10\},$ and $\{7,9\},$ and there are $8$ choices for $\{c,d\},$ namely $\{4,20\},\,\{5,19\},\,\{6,18\},\,\{7,17\},\,\{8,16\},\,\{9,15\},\{10,14\},$ and $\{11,13\}.$ Thus a four-element subset of the first type can be formed by taking the union of one of $7$ two-element subsets with one of $8$ two-element subsets as long as these two subsets are disjoint. There are $10$ such pairings that are not disjoint out of the $7\cdot 8=56$ pairings, so there are $56-10=46$ subsets of the first type.

For subsets of the second type, there are $10$ choices for the value of $a$ $(4,\,5,\,6,\,7,\,9,\,10,\,11,\,13,\,14,\,15)$ such that $b=16-a$ and $c=24-a$ can be two other elements of the subset. Note that in each of these cases, $c-b=(24-a)-(16-a)=8.$ For each of these, there are $20-3=17$ other values that can be chosen for the element $d$ in the subset. But $10\cdot 17=170$ counts some subsets more than once. In particular, a subset is counted twice if $b+d=24$ or $c+d=16$. In such cases either $d=a+8$ or $d=a-8$. There are exactly $6$ subsets where the role of $a$ can be played by two different elements of the set. They are $\{1,7,9,15\},\,\{2,6,10,14\},\,\{3,5,11,13\},\,\{5,11,13,19\},\,\{6,10,14,18\},$ and $\{7,9,15,17\}$. Thus there are $170-6=164$ subsets of the second type.

In all, there are $46+164=210$ subsets with the required property.

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