Difference between revisions of "2017 AIME I Problems/Problem 11"
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WLOG let <math>m=4</math> | WLOG let <math>m=4</math> | ||
− | There is a <math>\frac{15}{28}</math> chance that exactly one of 1, 2, 3 is in the same row. | + | 1. There is a <math>\frac{15}{28}</math> chance that exactly one of 1, 2, 3 is in the same row with 4. |
− | There is a <math>\frac{2}{5}</math> chance that the other two smaller numbers end up in the same row. | + | There are 3 ways to select which of the smaller numbers will get in the row, and then 5 |
+ | |||
+ | ways to select the number larger than 4. | ||
+ | |||
+ | <math>\frac{\dbinom{3}{1}\cdot\dbinom{5}{1}}{\dbinom{8}{2}} = \frac{15}{28}</math> | ||
+ | |||
+ | 2. There is a <math>\frac{2}{5}</math> chance that the other two smaller numbers end up in the same row. | ||
+ | |||
+ | There are 2 ways to select the row that the two smaller number are in, and then <math>\dbinom{3}{2}</math> ways | ||
+ | |||
+ | to place the smaller numbers in the row. | ||
+ | |||
+ | <math>\frac{\dbinom{2}{1}\cdot\dbinom{3}{2}}{\dbinom{6}{2}} = \frac{2}{5}</math> | ||
<math>9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}</math>. | <math>9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}</math>. | ||
− | |||
==Solution 3== | ==Solution 3== |
Revision as of 11:05, 20 August 2019
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
, which 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
1. There is a chance that exactly one of 1, 2, 3 is in the same row with 4.
There are 3 ways to select which of the smaller numbers will get in the row, and then 5
ways to select the number larger than 4.
2. There is a chance that the other two smaller numbers end up in the same row.
There are 2 ways to select the row that the two smaller number are in, and then ways
to place the smaller numbers in the 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 in
ways. There are
ways to permute the numbers in the individual Rows.
Hence, our answer is
Solution 4
We see that if one of the medians is 5, then there are two remaining numbers greater than 5 and two less than 5, so it follows that . There are 3 ways to choose which row to have 5 in,
ways to choose the other two numbers in that row,
ways to arrange the numbers in that row, and
ways for the remaining numbers, for our answer is
.
-Stormersyle
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.