# Difference between revisions of "2006 AIME A Problems/Problem 4"

(→Solution) |
|||

Line 7: | Line 7: | ||

== Solution == | == Solution == | ||

+ | |||

+ | 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.