Difference between revisions of "2006 AIME II Problems/Problem 4"
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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}=462</math> such ordered 12-tuples. | ||
+ | |||
+ | == Solution 2 == | ||
+ | There are <math>\binom{12}{6}</math> ways to choose 6 numbers from <math> (1,2,3,\ldots,12) </math>, and then there will only be one way to order them. And since that <math>a_6<a_7</math>, only half the combinations will work, so the answer is <math>\frac{\{binom{12}{5}}{2}=462</math> 12-tuples - mathleticguyyy | ||
== See also == | == See also == |
Revision as of 22:19, 8 January 2019
Contents
[hide]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.
Solution 2
There are ways to choose 6 numbers from , and then there will only be one way to order them. And since that , only half the combinations will work, so the answer is 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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