Difference between revisions of "2017 AIME I Problems/Problem 11"
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==Solution 3== | ==Solution 3== | ||
We will make sure to multiply by <math>3!</math> in the end to account for all the possible permutation of the rows. | We will make sure to multiply by <math>3!</math> in the end to account for all the possible permutation of the rows. | ||
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WLOG, let <math>5</math> be present in the Row #<math>1</math>. | WLOG, let <math>5</math> be present in the Row #<math>1</math>. | ||
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Notice that <math>5</math> MUST be placed with a number lower than it and a number higher than it. | Notice that <math>5</math> MUST be placed with a number lower than it and a number higher than it. | ||
| − | This happens in <math>4 | + | |
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| + | This happens in <math>4\cdot4</math> ways. You can permutate Row #<math>1</math> in <math>3!</math> ways. | ||
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Now, take a look at Row <math>2</math> and Row <math>3</math>. | Now, take a look at Row <math>2</math> and Row <math>3</math>. | ||
| − | Because there are <math>6</math> numbers to choose from now, you can assign #'s to Row's #<math>2&3</math> in <math>\frac{\binom{6}{3}\cdot\binom{3}{3}}{2}</math> ways. There are <math>3!\cdot3!</math> ways to permutate the numbers in the individual Rows. | + | |
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| + | Because there are <math>6</math> numbers to choose from now, you can assign #'s to Row's #<math>2&3</math> in | ||
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| + | <math>\frac{\binom{6}{3}\cdot\binom{3}{3}}{2}</math> ways. There are <math>3!\cdot3!</math> ways to permutate the numbers in the individual Rows. | ||
Hence, our answer is <math>3!(4\cdot4\cdot3!\cdot{10}\cdot{3!}\cdot{3!}=3!(16\cdot60\cdot36)=3!(34560)\implies{207\boxed{360}}</math> | Hence, our answer is <math>3!(4\cdot4\cdot3!\cdot{10}\cdot{3!}\cdot{3!}=3!(16\cdot60\cdot36)=3!(34560)\implies{207\boxed{360}}</math> | ||
Revision as of 13:31, 14 October 2017
Problem 11
Consider arrangements of the 
numbers 
in a 
array. For each such arrangement, let 
, 
, and 
be the medians of the numbers in rows 
, 
, and 
respectively, and let 
be the median of 
. Let 
be the number of arrangements for which 
. Find the remainder when is divided by 
.
Solution 1
We know that if 
is a median, then will be the median of the medians.
WLOG, assume is in the upper left corner. One of the two other values in the top row needs to be below , and the other needs to be above . This can be done in 
ways.
The other 
can be arranged in 
ways.
Finally, accounting for when is in every other space, our answer is 
. But we only need the last digits, so 
is our answer.
~Solution by SuperSaiyanOver9000, mathics42
Solution 2
(Complementary Counting with probability)
Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric.
WLOG let 
There is a 
chance that exactly one of 1, 2, 3 is in the same row.
There is a 
chance that the other two smaller numbers end up in the same row.
.
Solution 3
We will make sure to multiply by 
in the end to account for all the possible permutation of the rows.
WLOG, let be present in the Row #.
Notice that MUST be placed with a number lower than it and a number higher than it.
This happens in 
ways. You can permutate Row # in ways.
Now, take a look at Row and Row .
Because there are numbers to choose from now, you can assign #'s to Row's #$2&3$ (Error compiling LaTeX. Unknown error_msg) in
ways. There are 
ways to permutate the numbers in the individual Rows.
Hence, our answer is 
See Also
| 2017 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
