Difference between revisions of "2006 AIME II Problems/Problem 4"
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Case 1: <math>a_1 = 12</math> | Case 1: <math>a_1 = 12</math> | ||
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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> | ||
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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. | ||
Revision as of 21:49, 27 October 2020
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 of the choices will work, so the answer is 12-tuples - mathleticguyyy
Solution 3
Clearly, , and either or is 12.
Case 1:
In this case, there are 4 empty spaces between and , and 6 empty spaces between and . 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:
In this case, there are 5 empty spaces between and , and 5 empty spaces between and . 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.
-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.