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 17:14, 16 July 2006
Problem
Let be a permutation of
for which
![$a_1>a_2>a_3>a_4>a_5>a_6 \mathrm{\ and \ } a_6<a_7<a_8<a_9<a_{10}<a_{11}<a_{12}.$](http://latex.artofproblemsolving.com/7/1/4/714f8aaced4000d779337cb9bddced3aff9ed3e7.png)
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.