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Difference between revisions of "2006 AIME II 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.
+
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}=\boxed{462}</math> such ordered 12-tuples.
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 15:30, 29 February 2020

Problem

Let $(a_1,a_2,a_3,\ldots,a_{12})$ be a permutation of $(1,2,3,\ldots,12)$ for which

$a_1>a_2>a_3>a_4>a_5>a_6 \mathrm{\ and \ } a_6<a_7<a_8<a_9<a_{10}<a_{11}<a_{12}.$

An example of such a permutation is $(6,5,4,3,2,1,7,8,9,10,11,12).$ Find the number of such permutations.

Solution

Clearly, $a_6=1$. Now, consider selecting $5$ of the remaining $11$ values. Sort these values in descending order, and sort the other $6$ values in ascending order. Now, let the $5$ selected values be $a_1$ through $a_5$, and let the remaining $6$ be $a_7$ through ${a_{12}}$. It is now clear that there is a bijection between the number of ways to select $5$ values from $11$ and ordered 12-tuples $(a_1,\ldots,a_{12})$. Thus, there will be ${11 \choose 5}=\boxed{462}$ such ordered 12-tuples.

Solution 2

There are $\binom{12}{6}$ ways to choose 6 numbers from $(1,2,3,\ldots,12)$, and then there will only be one way to order them. And since that $a_6<a_7$, only half of the choices will work, so the answer is $\frac{\binom{12}{6}}{2}=462$ 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

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