Difference between revisions of "2006 AIME A Problems/Problem 4"
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== Solution == | == Solution == | ||
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| + | Clearly, <math>a_6=1</math>. Now, consider selecting <math>5</math> of the remaining <math>11</math> values. Sort these values in descending order, and sort the other <math>6</math> values in ascending order. Now, let the <math>5</math> selected values be <math>a_1</math> through <math>a_5</math>, and let the remaining <math>6</math> be <math>a_7</math> through <math>{a_{12}}</math>. It is now clear that there is a bijection between the number of ways to select <math>5</math> values from <math>11</math> and ordered 12-tuples <math>(a_1,\ldots,a_{12})</math>. Thus, there will be <math>{11 \choose 5}=462</math> such ordered 12-tuples. | ||
== See also == | == See also == | ||
Revision as of 16:14, 16 July 2006
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.
