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

Revision as of 18:21, 14 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

The integers $x \in \{8, \dots , 14 \}$ must each be the larger elements of a distinct pair. So, we 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 more constraints, so there are $\mathbf{4!} \textbf{ ways}$ to choose partners for $\{1,2,3,4\}$ .

The answer is $1 \cdot 3! \cdot 4! = \boxed{\textbf{(E) } 144}$ .

~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
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

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

@@ Line 31: / Line 31: @@
 ==Solution 3==
-The integers <math>x \in \{8, \dots , 14 \}</math> must each be the larger elements of a distinct pair.
+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>:
-Assign partners in decreasing order for <math>x \in \{7, \dots, 1\}</math>:
 Note that <math>7</math> must pair with <math>14</math>: <math>\mathbf{1} \textbf{ choice}</math>.

Page

Toolbox

Search