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Difference between revisions of "2022 AMC 10A Problems/Problem 14"

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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 (Casework)==
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==Solution 1 (Multiplication Principle)==
  
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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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Clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs, so are the integers from <math>1</math> through <math>7.</math> Note that <math>7</math> must pair with <math>14.</math>
  
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Note that <math>6</math> can pair with either <math>12</math> or <math>13.</math> From here, we consider casework:
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We pair the numbers <math>1,2,3,4,5,6</math> with the numbers <math>8,9,10,11,12,13</math> systematically:
  
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* If <math>6</math> pairs with <math>12,</math> then <math>5</math> can pair with one of <math>10,11,13.</math> After that, each of <math>1,2,3,4</math> does not have any restrictions. This case produces <math>3\cdot4!=72</math> ways.
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* <math>6</math> can pair with either <math>12</math> or <math>13.</math>
  
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* If <math>6</math> pairs with <math>13,</math> then <math>5</math> can pair with one of <math>10,11,12.</math> After that, each of <math>1,2,3,4</math> does not have any restrictions. This case produces <math>3\cdot4!=72</math> ways.
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* <math>5</math> can pair with any of the three remaining numbers from <math>10,11,12,13.</math>
  
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Together, the answer is <math>72+72=\boxed{\textbf{(E) } 144}.</math>
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* <math>1,2,3,4</math> can pair with the other four remaining numbers from <math>8,9,10,11,12,13</math> without restrictions.
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Together, the answer is <math>2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
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You forgot to mention that <math>1,2,3,4</math> cannot be paired with each other because the "balance" between the larger numbers and smaller numbers will be lost. In other words, if this happens, one of the pairs must be something like <math>(10,11)</math> which is illegal.
 
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-Fireball9746
 
  
 
==Solution 2 (Multiplication Principle)==
 
==Solution 2 (Multiplication Principle)==

Latest revision as of 23:52, 23 October 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 (Multiplication Principle)

Clearly, the integers from $8$ through $14$ must be in different pairs, so are the integers from $1$ through $7.$ Note that $7$ must pair with $14.$

We pair the numbers $1,2,3,4,5,6$ with the numbers $8,9,10,11,12,13$ systematically:

  • $6$ can pair with either $12$ or $13.$
  • $5$ can pair with any of the three remaining numbers from $10,11,12,13.$
  • $1,2,3,4$ can pair with the other four remaining numbers from $8,9,10,11,12,13$ without restrictions.

Together, the answer is $2\cdot3\cdot4!=\boxed{\textbf{(E) } 144}.$

~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 (RMM club)

https://youtu.be/DwCE1wu5hrA

Video Solution by Lucas637 (Fast and Easy)

https://www.youtube.com/watch?v=egQK11g54mA

Video Solution by TheBeautyofMath

https://youtu.be/0kkc4-y8TkU?t=1367

~IceMatrix

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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