Difference between revisions of "2017 AIME I Problems/Problem 11"
(Created page with "We know that if <math>5</math> is a median, then <math>5</math> will be the median of the medians. WLOG, assume <math>5</math> is in the upper left corner. One of the two oth...") |
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~Solution by SuperSaiyanOver9000, mathics42 | ~Solution by SuperSaiyanOver9000, mathics42 | ||
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| + | Solution 2 (Complementary Counting with probability) | ||
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| + | Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric. | ||
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| + | WLOG let <math>m=4</math> | ||
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| + | There is a <math>\frac{15}{28}</math> chance that exactly one of 1, 2, 3 is in the same row. | ||
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| + | There is a <math>\frac{2}{5}</math> chance that the other two smaller numbers end up in the same row. | ||
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| + | <math>9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}</math>. | ||
Revision as of 17:41, 8 March 2017
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.
.
