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Difference between revisions of "2022 AMC 10A Problems/Problem 14"
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{{duplicate|[[2022 AMC 10A Problems/Problem 14|2022 AMC 10A #14]] and [[2022 AMC 12A Problems/Problem 10|2022 AMC 12A #10]]}}
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==Problem==
 
==Problem==
  
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<math>\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144</math>
 
<math>\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144</math>
  
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==Solution 1==
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==Solution 1 (Casework)==
  
 
Clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs, and <math>7</math> must pair with <math>14.</math>
 
Clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs, and <math>7</math> must pair with <math>14.</math>
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2==
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==Solution 2 (Multiplication Principle)==
  
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As said above, clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs.
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As said in Solution 1, clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs.
  
 
We know that <math>8</math> or <math>9</math> can pair with any integer from <math>1</math> to <math>4</math>, <math>10</math> or <math>11</math> can pair with any integer from <math>1</math> to <math>5</math>, and <math>12</math> or <math>13</math> can pair with any integer from <math>1</math> to <math>6</math>. Thus, <math>8</math> will have <math>4</math> choices to pair with, <math>9</math> will then have <math>3</math> choices to pair with (<math>9</math> cannot pair with the same number as the one <math>8</math> pairs with). <math>10</math> cannot pair with the numbers <math>8</math> and <math>9</math> has paired with but can also now pair with <math>5</math>, so there are <math>3</math> choices. <math>11</math> cannot pair with <math>8</math>'s, <math>9</math>'s, or <math>10</math>'s paired numbers, so there will be <math>2</math> choices for <math>11</math>. <math>12</math> can pair with an integer from <math>1</math> to <math>5</math> that hasn't been paired with already, or it can pair with <math>6</math>. <math>13</math> will only have one choice left, and <math>7</math> must pair with <math>14</math>.
 
We know that <math>8</math> or <math>9</math> can pair with any integer from <math>1</math> to <math>4</math>, <math>10</math> or <math>11</math> can pair with any integer from <math>1</math> to <math>5</math>, and <math>12</math> or <math>13</math> can pair with any integer from <math>1</math> to <math>6</math>. Thus, <math>8</math> will have <math>4</math> choices to pair with, <math>9</math> will then have <math>3</math> choices to pair with (<math>9</math> cannot pair with the same number as the one <math>8</math> pairs with). <math>10</math> cannot pair with the numbers <math>8</math> and <math>9</math> has paired with but can also now pair with <math>5</math>, so there are <math>3</math> choices. <math>11</math> cannot pair with <math>8</math>'s, <math>9</math>'s, or <math>10</math>'s paired numbers, so there will be <math>2</math> choices for <math>11</math>. <math>12</math> can pair with an integer from <math>1</math> to <math>5</math> that hasn't been paired with already, or it can pair with <math>6</math>. <math>13</math> will only have one choice left, and <math>7</math> must pair with <math>14</math>.
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~Scarletsyc
 
~Scarletsyc
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==Solution 3 (Generalization)==
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The integers <math>x \in \{8, \ldots , 14 \}</math> must each be the larger elements of a distinct pair.
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Assign partners in decreasing order for <math>x \in \{7, \dots, 1\}</math>:
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Note that <math>7</math> must pair with <math>14</math>: <math>\mathbf{1} \textbf{ choice}</math>.
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For <math>5 \leq x \leq 7</math>, the choices are <math>\{2x, \dots, 14\} - \{ \text{previous choices}\}</math>. As <math>x</math> decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding <math>\mathbf{3!} \textbf{ combined choices}</math>.
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After assigning a partner to <math>5</math>, there are no invalid pairings for yet-unpaired numbers, so there are <math>\mathbf{4!} \textbf{ ways}</math> to choose partners for <math>\{1,2,3,4\}</math>.
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The answer is <math>3! \cdot 4! = \boxed{\textbf{(E) } 144}</math>.
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In general, for <math>1,\ldots,2n</math>, the same logic yields answer: <math>\left\lfloor\dfrac{n}{2}\right\rfloor! \cdot \left\lceil\dfrac{n}{2}\right\rceil!</math>
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~oinava
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==Video Solution by Education, the Study of Everything==
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https://youtu.be/k6EUl65wS9Q
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== Video Solution by Sohil Rathi==
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https://youtu.be/V1jOj8ysd_w
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~ pi_is_3.14
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==Video Solution (Smart and Simple)==
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https://youtu.be/7yAh4MtJ8a8?si=jyIdy-jZb2raj3cM&t=1800
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 +
~Math-X
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==Video Solution==
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https://youtu.be/DwCE1wu5hrA
  
 
== See Also ==
 
== See Also ==

Latest revision as of 20:59, 21 February 2024

The following problem is from both the 2022 AMC 10A #14 and 2022 AMC 12A #10, so both problems redirect to this page.

Problem

How many ways are there to split the integers $1$ through $14$ into $7$ pairs such that in each pair, the greater number is at least $2$ times the lesser number?

$\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144$

Solution 1 (Casework)

Clearly, the integers from $8$ through $14$ must be in different pairs, and $7$ must pair with $14.$

Note that $6$ can pair with either $12$ or $13.$ From here, we consider casework:

  • If $6$ pairs with $12,$ then $5$ can pair with one of $10,11,13.$ After that, each of $1,2,3,4$ does not have any restrictions. This case produces $3\cdot4!=72$ ways.
  • If $6$ pairs with $13,$ then $5$ can pair with one of $10,11,12.$ After that, each of $1,2,3,4$ does not have any restrictions. This case produces $3\cdot4!=72$ ways.

Together, the answer is $72+72=\boxed{\textbf{(E) } 144}.$

~MRENTHUSIASM

Solution 2 (Multiplication Principle)

As said in Solution 1, clearly, the integers from $8$ through $14$ must be in different pairs.

We know that $8$ or $9$ can pair with any integer from $1$ to $4$, $10$ or $11$ can pair with any integer from $1$ to $5$, and $12$ or $13$ can pair with any integer from $1$ to $6$. Thus, $8$ will have $4$ choices to pair with, $9$ will then have $3$ choices to pair with ($9$ cannot pair with the same number as the one $8$ pairs with). $10$ cannot pair with the numbers $8$ and $9$ has paired with but can also now pair with $5$, so there are $3$ choices. $11$ cannot pair with $8$'s, $9$'s, or $10$'s paired numbers, so there will be $2$ choices for $11$. $12$ can pair with an integer from $1$ to $5$ that hasn't been paired with already, or it can pair with $6$. $13$ will only have one choice left, and $7$ must pair with $14$.

So, the answer is $4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{\textbf{(E) } 144}.$

~Scarletsyc

Solution 3 (Generalization)

The integers $x \in \{8, \ldots , 14 \}$ must each be the larger elements of a distinct pair.

Assign partners in decreasing order for $x \in \{7, \dots, 1\}$:

Note that $7$ must pair with $14$: $\mathbf{1} \textbf{ choice}$.

For $5 \leq x \leq 7$, the choices are $\{2x, \dots, 14\} - \{ \text{previous choices}\}$. As $x$ decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding $\mathbf{3!} \textbf{ combined choices}$.

After assigning a partner to $5$, there are no invalid pairings for yet-unpaired numbers, so there are $\mathbf{4!} \textbf{ ways}$ to choose partners for $\{1,2,3,4\}$.

The answer is $3! \cdot 4! = \boxed{\textbf{(E) } 144}$.

In general, for $1,\ldots,2n$, the same logic yields answer: $\left\lfloor\dfrac{n}{2}\right\rfloor! \cdot \left\lceil\dfrac{n}{2}\right\rceil!$

~oinava

Video Solution by Education, the Study of Everything

https://youtu.be/k6EUl65wS9Q

Video Solution by Sohil Rathi

https://youtu.be/V1jOj8ysd_w

~ pi_is_3.14

Video Solution (Smart and Simple)

https://youtu.be/7yAh4MtJ8a8?si=jyIdy-jZb2raj3cM&t=1800

~Math-X

Video Solution

https://youtu.be/DwCE1wu5hrA

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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