Difference between revisions of "2022 AMC 10A Problems/Problem 14"
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After assigning a partner to <math>5</math>, there are no more constraints, 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 more constraints, so there are <math>\mathbf{4!}</math> ways to choose partners for <math>\{1,2,3,4\}</math>. | ||
| − | 1 \cdot <math>3! \cdot 4! = \boxed{144} | + | <math>1 \cdot </math>3! \cdot 4! = \boxed{144}$. |
~oinava | ~oinava | ||
Revision as of 15:24, 14 March 2023
Contents
Problem
How many ways are there to split the integers 
through 
into 
pairs such that in each pair, the greater number is at least 
times the lesser number?
Solution 1
Clearly, the integers from 
through must be in different pairs, and must pair with 
Note that 
can pair with either 
or 
From here, we consider casework:
- If
pairs with 
then 
can pair with one of 
After that, each of 
does not have any restrictions. This case produces 
ways.
then 
After that, each of 
does not have any restrictions. This case produces 
ways.- If pairs with
then can pair with one of 
After that, each of does not have any restrictions. This case produces ways.
then 
After that, each of Together, the answer is 
~MRENTHUSIASM
Solution 2
As said above, clearly, the integers from through must be in different pairs.
We know that or 
can pair with any integer from to 
, 
or 
can pair with any integer from to 
, and or 
can pair with any integer from to . Thus, will have choices to pair with, will then have 
choices to pair with ( cannot pair with the same number as the one pairs with). cannot pair with the numbers and has paired with but can also now pair with , so there are choices. cannot pair with 's, 's, or 's paired numbers, so there will be choices for . can pair with an integer from to that hasn't been paired with already, or it can pair with . will only have one choice left, and must pair with .
So, the answer is 
~Scarletsyc
Solution 3 (Most contrained first)
The integers 
must each be the larger elements of a distinct pair.
Assign partners in decreasing order for 
:
must pair with : 
.
For 
, the choices are 
. As 
decreases by 1, The minuend increases by 2 elements, and the subtrahend increases by 1 element, so the difference increases by 1, yielding 
.
After assigning a partner to , there are no more constraints, so there are 
ways to choose partners for 
.
3! \cdot 4! = \boxed{144}$.
~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. 
