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.

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

~ pi_is_3.14

~Math-X