Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 3)
(Titled all solutions to make them consistent. Also, Sentences should never begin with symbols.)
Line 5: Line 5:
 
<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>
  
−
==Solution 1==
+
==Solution 1 (Casework)==
  
 
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>
 
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>
Line 19: Line 19:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
==Solution 2==
+
==Solution 2 (Multiplication Principle)==
  
−
As said above, clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs.
+
As said in Solution 1, clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs.
  
 
We know that <math>8</math> or <math>9</math> can pair with any integer from <math>1</math> to <math>4</math>, <math>10</math> or <math>11</math> can pair with any integer from <math>1</math> to <math>5</math>, and <math>12</math> or <math>13</math> can pair with any integer from <math>1</math> to <math>6</math>. Thus, <math>8</math> will have <math>4</math> choices to pair with, <math>9</math> will then have <math>3</math> choices to pair with (<math>9</math> cannot pair with the same number as the one <math>8</math> pairs with). <math>10</math> cannot pair with the numbers <math>8</math> and <math>9</math> has paired with but can also now pair with <math>5</math>, so there are <math>3</math> choices. <math>11</math> cannot pair with <math>8</math>'s, <math>9</math>'s, or <math>10</math>'s paired numbers, so there will be <math>2</math> choices for <math>11</math>. <math>12</math> can pair with an integer from <math>1</math> to <math>5</math> that hasn't been paired with already, or it can pair with <math>6</math>. <math>13</math> will only have one choice left, and <math>7</math> must pair with <math>14</math>.
 
We know that <math>8</math> or <math>9</math> can pair with any integer from <math>1</math> to <math>4</math>, <math>10</math> or <math>11</math> can pair with any integer from <math>1</math> to <math>5</math>, and <math>12</math> or <math>13</math> can pair with any integer from <math>1</math> to <math>6</math>. Thus, <math>8</math> will have <math>4</math> choices to pair with, <math>9</math> will then have <math>3</math> choices to pair with (<math>9</math> cannot pair with the same number as the one <math>8</math> pairs with). <math>10</math> cannot pair with the numbers <math>8</math> and <math>9</math> has paired with but can also now pair with <math>5</math>, so there are <math>3</math> choices. <math>11</math> cannot pair with <math>8</math>'s, <math>9</math>'s, or <math>10</math>'s paired numbers, so there will be <math>2</math> choices for <math>11</math>. <math>12</math> can pair with an integer from <math>1</math> to <math>5</math> that hasn't been paired with already, or it can pair with <math>6</math>. <math>13</math> will only have one choice left, and <math>7</math> must pair with <math>14</math>.
Line 29: Line 29:
 
~Scarletsyc
 
~Scarletsyc
  
−
==Solution 3 (Most contrained first, generalized)==
+
==Solution 3 (Generalization)==
  
−
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, \ldots , 14 \}</math> must each be the larger elements of a distinct pair.
  
 
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>:
  
−
<math>7</math> must pair with <math>14</math>: <math>\mathbf{1} \textbf{ choice}</math>.
+
Note that <math>7</math> must pair with <math>14</math>: <math>\mathbf{1} \textbf{ choice}</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>.
 
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>.
  
−
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>.
+
After assigning a partner to <math>5</math>, there are no invalid pairings for yet-unpaired numbers, so there are <math>\mathbf{4!} \textbf{ ways}</math> to choose partners for <math>\{1,2,3,4\}</math>.
  
−
<math>3! \cdot 4! = \boxed{144}</math>.
+
The answer is <math>3! \cdot 4! = \boxed{\textbf{(E) } 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>
+
In general, for <math>1,\ldots,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
 
~oinava

Revision as of 22:41, 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 (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

Video Solution by OmegaLearn

https://youtu.be/V1jOj8ysd_w

~ pi_is_3.14

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_14&oldid=191015"