Difference between revisions of "2022 AMC 10A Problems/Problem 14"
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<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> | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 2== | + | ==Solution 2 (Multiplication Principle)== |
− | As said | + | 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>. | ||
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~Scarletsyc | ~Scarletsyc | ||
− | ==Solution 3 ( | + | ==Solution 3 (Generalization)== |
− | The integers <math>x \in \{8, \ | + | 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> | + | 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, | + | 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 23:41, 15 March 2023
Contents
Problem
How many ways are there to split the integers through
into
pairs such that in each pair, the greater number is at least
times the lesser number?
Solution 1 (Casework)
Clearly, the integers from through
must be in different pairs, and
must pair with
Note that can pair with either
or
From here, we consider casework:
- If
pairs with
then
can pair with one of
After that, each of
does not have any restrictions. This case produces
ways.
- If
pairs with
then
can pair with one of
After that, each of
does not have any restrictions. This case produces
ways.
Together, the answer is
~MRENTHUSIASM
Solution 2 (Multiplication Principle)
As said in Solution 1, clearly, the integers from through
must be in different pairs.
We know that or
can pair with any integer from
to
,
or
can pair with any integer from
to
, and
or
can pair with any integer from
to
. Thus,
will have
choices to pair with,
will then have
choices to pair with (
cannot pair with the same number as the one
pairs with).
cannot pair with the numbers
and
has paired with but can also now pair with
, so there are
choices.
cannot pair with
's,
's, or
's paired numbers, so there will be
choices for
.
can pair with an integer from
to
that hasn't been paired with already, or it can pair with
.
will only have one choice left, and
must pair with
.
So, the answer is
~Scarletsyc
Solution 3 (Generalization)
The integers must each be the larger elements of a distinct pair.
Assign partners in decreasing order for :
Note that must pair with
:
.
For , the choices are
. As
decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding
.
After assigning a partner to , there are no invalid pairings for yet-unpaired numbers, so there are
to choose partners for
.
The answer is .
In general, for , the same logic yields answer:
~oinava
Video Solution by OmegaLearn
~ 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.