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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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https://youtu.be/7yAh4MtJ8a8?si=jyIdy-jZb2raj3cM&t=1800
 
https://youtu.be/7yAh4MtJ8a8?si=jyIdy-jZb2raj3cM&t=1800
  
−
~ Math-X
+
~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 ==
 
== See Also ==

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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