2022 AMC 10A Problems/Problem 14

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

1 Problem
2 Solution 1 (Multiplication Principle)
3 Solution 2 (Multiplication Principle)
4 Solution 3 (Generalization)
5 Video Solution by Education, the Study of Everything
6 Video Solution by Sohil Rathi
7 Video Solution (Smart and Simple)
8 Video Solution (RMM club)
9 Video Solution by Lucas637 (Fast and Easy)
10 Video Solution by TheBeautyofMath
11 See Also

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, and $7$ 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

Note: You could make an argument that 2 can pair with 4 (or 1 with 3, 1 with 2, etc.)instead of the biggest 4 remaining numbers and so we have undercounted. H

owever, this is false because if you pair two numbers with each other in the smaller subset, that would leave 2 numbers in the smaller subset ${1,2,3,4}$ and 4 numbers from the bigger subset ${8,9,10,11,12,13}$ left to pair. However that would require two numbers from the bigger subset to pair together, which can never work.

~mathboy282

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

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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