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

Revision as of 13:30, 15 March 2023

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

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

As said above, 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 (Most contrained first, generalized)

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

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

$7$ 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 $\mathbf{4!}$ ways to choose partners for $\{1,2,3,4\}$ .

$3! \cdot 4! = \boxed{144}$ .

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

~oinava

Video Solution by OmegaLearn

https://youtu.be/V1jOj8ysd_w

~ pi_is_3.14

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 10 Problems and Solutions

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 29: / Line 29: @@
 ~Scarletsyc
-==Solution 3==
+==Solution 3 (Most contrained first, generalized)==
-The integers <math>x \in \{8, \dots , 14 \}</math> must each be the larger elements of a distinct pair. So, we assign partners in decreasing order for <math>x \in \{7, \dots, 1\}</math>:
+The integers <math>x \in \{8, \dots , 14 \}</math> must each be the larger elements of a distinct pair.
-Note that <math>7</math> must pair with <math>14</math>: <math>\mathbf{1} \textbf{ choice}</math>.
+Assign partners in decreasing order for <math>x \in \{7, \dots, 1\}</math>:
-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 <math>1</math>, The minuend increases by <math>2</math> elements, and the subtrahend increases by <math>1</math> element, so the difference increases by <math>1</math>, yielding <math>\mathbf{3!} \textbf{ combined choices}</math>.
+<math>7</math> must pair with <math>14</math>: <math>\mathbf{1} \textbf{ choice}</math>.
-After assigning a partner to <math>5</math>, there are no more constraints, so there are <math>\mathbf{4!} \textbf{ ways}</math> to choose partners for <math>\{1,2,3,4\}</math>.
+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>.
-The answer is <math>1 \cdot 3! \cdot 4! = \boxed{\textbf{(E) } 144}</math>.
+After assigning a partner to <math>5</math>, there are no invalid pairings for yet-unpaired numbers, so there are <math>\mathbf{4!}</math> ways to choose partners for <math>\{1,2,3,4\}</math>.
+<math>3! \cdot 4! = \boxed{144}</math>.
+In general, for <math>1,...,2n</math>, the same logic yields answer: <math>\left\lfloor\dfrac{n}{2}\right\rfloor! \cdot \left\lceil\dfrac{n}{2}\right\rceil!</math>
 ~oinava

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