Difference between revisions of "2006 AIME A Problems/Problem 4"
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== Problem == | == Problem == | ||
| − | Let <math> (a_1,a_2,a_3,\ldots,a_{12}) </math> be a permutation of <math> (1,2,3,\ldots,12) </math> for which | + | Let <math> (a_1,a_2,a_3,\ldots,a_{12}) </math> be a [[permutation]] of <math> (1,2,3,\ldots,12) </math> for which |
| − | <center><math> 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}. </math></ | + | <div style="text-align:center;"><math> 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}. </math></div> |
An example of such a permutation is <math> (6,5,4,3,2,1,7,8,9,10,11,12). </math> Find the number of such permutations. | An example of such a permutation is <math> (6,5,4,3,2,1,7,8,9,10,11,12). </math> Find the number of such permutations. | ||
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== 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. | + | 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 == | ||
| − | + | {{AIME box|year=2006|n=II|num-b=3|num-a=5}} | |
[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 15:57, 12 March 2007
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.
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 | ||
