Difference between revisions of "2006 AIME II Problems/Problem 4"
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== Solution 2 == | == Solution 2 == | ||
| − | There are <math>\binom{12}{6}</math> ways to choose 6 numbers from <math> (1,2,3,\ldots,12) </math>, and then there will only be one way to order them. And since that <math>a_6<a_7</math>, only half the combinations will work, so the answer is <math>\frac | + | There are <math>\binom{12}{6}</math> ways to choose 6 numbers from <math> (1,2,3,\ldots,12) </math>, and then there will only be one way to order them. And since that <math>a_6<a_7</math>, only half the combinations will work, so the answer is <math>\frac{\binom{12}{6}}{2}=462</math> 12-tuples - mathleticguyyy |
== See also == | == See also == | ||
Revision as of 22:53, 9 January 2019
Contents
Problem
Let 
be a permutation of 
for which
An example of such a permutation is 
Find the number of such permutations.
Solution
Clearly, 
. Now, consider selecting 
of the remaining 
values. Sort these values in descending order, and sort the other 
values in ascending order. Now, let the selected values be 
through 
, and let the remaining be 
through 
. It is now clear that there is a bijection between the number of ways to select values from and ordered 12-tuples 
. Thus, there will be 
such ordered 12-tuples.
Solution 2
There are 
ways to choose 6 numbers from , and then there will only be one way to order them. And since that 
, only half the combinations will work, so the answer is 
12-tuples - mathleticguyyy
See also
| 2006 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
