Difference between revisions of "2022 AMC 10A Problems/Problem 14"
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~Scarletsyc | ~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. | + | The integers <math>x \in \{8, \dots , 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>: | |
− | + | <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>. | |
− | + | 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 | ~oinava |
Revision as of 13:30, 15 March 2023
Contents
[hide]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
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
As said above, 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 (Most contrained first, generalized)
The integers must each be the larger elements of a distinct pair.
Assign partners in decreasing order for :
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 ways to choose partners for .
.
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 |
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