Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2006 AIME II Problems/Problem 4"

(→‎Solution 3)
m (Undo revision 215042 by Makemake (talk))
(Tag: Undo)
 
(6 intermediate revisions by 3 users not shown)
Line 16: Line 16:
  
 
==Solution 3==
 
==Solution 3==
−
Clearly, <math>a_6=1</math>, and either <math>a_1</math> or <math>a_12</math> is 12.
+
Clearly, <math>a_6=1</math>, and either <math>a_1</math> or <math>a_{12}</math> is 12.
  
  
 
Case 1: <math>a_1 = 12</math>
 
Case 1: <math>a_1 = 12</math>
  
−
In this case, there are 4 empty spaces between <math>a_1</math> and <math>a_6</math>, and 6 empty spaces between <math>a_6</math> and <math>a_12</math>. <math>\binom{10}{4}</math> is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.
+
In this case, there are 4 empty spaces between <math>a_1</math> and <math>a_6</math>, and 6 empty spaces between <math>a_6</math> and <math>a_{12}</math>. <math>\binom{10}{4}</math> is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.
  
  
−
Case 2: <math>a_12 = 12</math>
+
Case 2: <math>a_{12} = 12</math>
  
−
In this case, there are 5 empty spaces between <math>a_1</math> and <math>a_6</math>, and 5 empty spaces between <math>a_6</math> and <math>a_12</math>. <math>\binom{10}{5}</math> is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.
+
In this case, there are 5 empty spaces between <math>a_1</math> and <math>a_6</math>, and 5 empty spaces between <math>a_6</math> and <math>a_{12}</math>. <math>\binom{10}{5}</math> is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.
  
−
<math>210 + 252 = \boxed{462}</math>
+
<math>210 + 252 = \boxed{462}</math> ordered 12-tuples.
  
  

Latest revision as of 13:12, 20 April 2024

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

Solution 3

Clearly, $a_6=1$, and either $a_1$ or $a_{12}$ is 12.


Case 1: $a_1 = 12$

In this case, there are 4 empty spaces between $a_1$ and $a_6$, and 6 empty spaces between $a_6$ and $a_{12}$. $\binom{10}{4}$ is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.


Case 2: $a_{12} = 12$

In this case, there are 5 empty spaces between $a_1$ and $a_6$, and 5 empty spaces between $a_6$ and $a_{12}$. $\binom{10}{5}$ is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.

$210 + 252 = \boxed{462}$ ordered 12-tuples.


-jackshi2006

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_4&oldid=218379"