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

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As said above, clearly, the integers from <math>8</math> through <math>14</math> must be in different pairs.
 
As said above, 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 a number 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>.
  
 
So, the answer is <math>4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{\textbf{(E) } 144}.</math>
 
So, the answer is <math>4\cdot3\cdot3\cdot2\cdot2\cdot1\cdot1=\boxed{\textbf{(E) } 144}.</math>

Revision as of 09:38, 12 November 2022

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

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

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