Difference between revisions of "2018 AIME II Problems/Problem 11"
(→Solution 4 (You won't get 458, 459, 460, 462, 465, 467, etc. with this method!!!)) |
(→Solution 4 (You won't get 458, 459, 460, 462, 465, 467, etc. with this method!!!)) |
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For every <math>i \in S</math>, we can divide <math>S</math> into two subsets: <cmath>S_{1\to i}=\{1,2,...i\}; S_{i+1\to n}=\{i+1,i+2,...,n\}</cmath> | For every <math>i \in S</math>, we can divide <math>S</math> into two subsets: <cmath>S_{1\to i}=\{1,2,...i\}; S_{i+1\to n}=\{i+1,i+2,...,n\}</cmath> | ||
− | Define permutation <math>P</math> as the permutation satisfy the condition of this problem. Then according to the condition of this problem, for each <math>i\in \{1,2,...,n-1\}</math>, <math>P</math> is not a permutation of set <math>S_{1\to i}</math>. For each <math>i\in \{1,2,...,n | + | Define permutation <math>P</math> as the permutation satisfy the condition of this problem. Then according to the condition of this problem, for each <math>i\in \{1,2,...,n-1\}</math>, <math>P</math> is not a permutation of set <math>S_{1\to i}</math>. For each <math>i\in \{1,2,...,n\}</math>, mark the number of permutation <math>P</math> of set <math>S</math> as <math>P_{k}</math>, where <math>k=i</math>, mark the number of permutation <math>P</math> for set <math>S_{i+1\to n}</math> as <math>P_{i}</math>; then, according to the condition of this problem, the permutation for <math>S_{i+1\to n}</math> is unrestricted, so the number of the unrestricted permutation of <math>S_{i+1\to n}</math> is <math>(n-i)!</math>. As a result, for each <math>i\in \{1,2,...,n\}</math>, the total number of permutation <math>P</math> is <cmath>P_{k}=P_{i}(n-i)!</cmath> |
Notice that according to the condition of this problem, if you sum all <math>P_{k}</math> up, you will get the total number of permutation of <math>S</math>, that is, <cmath>N=\sum^{n}_{k=1}{P_{k}}=\sum^{n}_{i=1}{P_{i}(n-i)!}=n!</cmath> | Notice that according to the condition of this problem, if you sum all <math>P_{k}</math> up, you will get the total number of permutation of <math>S</math>, that is, <cmath>N=\sum^{n}_{k=1}{P_{k}}=\sum^{n}_{i=1}{P_{i}(n-i)!}=n!</cmath> | ||
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<cmath>P_{5}=71</cmath> | <cmath>P_{5}=71</cmath> | ||
− | <cmath>P_{6}= | + | <cmath>P_{6}=461</cmath> |
+ | |||
+ | So the total number of permutations satisify this problem is <math>P_{6}=boxed{461}</math>. | ||
~Solution by <math>BladeRunnerAUG</math> (Frank FYC) | ~Solution by <math>BladeRunnerAUG</math> (Frank FYC) |
Revision as of 10:41, 6 October 2018
Contents
Problem
Find the number of permutations of such that for each with , at least one of the first terms of the permutation is greater than .
Solution 1
If the first number is , then there are no restrictions. There are , or ways to place the other numbers.
If the first number is , can go in four places, and there are ways to place the other numbers. ways.
If the first number is , ....
4 6 _ _ _ _ 24 ways
4 _ 6 _ _ _ 24 ways
4 _ _ 6 _ _ 24 ways
4 _ _ _ 6 _ 5 must go between and , so there are ways.
ways if 4 is first.
If the first number is , ....
3 6 _ _ _ _ 24 ways
3 _ 6 _ _ _ 24 ways
3 1 _ 6 _ _ 4 ways
3 2 _ 6 _ _ 4 ways
3 4 _ 6 _ _ 6 ways
3 5 _ 6 _ _ 6 ways
3 5 _ _ 6 _ 6 ways
3 _ 5 _ 6 _ 6 ways
3 _ _ 5 6 _ 4 ways
ways
If the first number is , ....
2 6 _ _ _ _ 24 ways
2 _ 6 _ _ _ 18 ways
2 3 _ 6 _ _ 4 ways
2 4 _ 6 _ _ 4 ways
2 4 _ 6 _ _ 6 ways
2 5 _ 6 _ _ 6 ways
2 5 _ _ 6 _ 6 ways
2 _ 5 _ 6 _ 4 ways
2 4 _ 5 6 _ 2 ways
2 3 4 5 6 1 1 way
ways
Grand Total :
Solution 2
If is the first number, then there are no restrictions. There are , or ways to place the other numbers.
If is the second number, then the first number can be or , and there are ways to place the other numbers. ways.
If is the third number, then we cannot have the following:
1 _ 6 _ _ _ 24 ways
2 1 6 _ _ _ 6 ways
ways
If is the fourth number, then we cannot have the following: 1 _ _ 6 _ _ 24 ways
2 1 _ 6 _ _ 6 ways
2 3 1 6 _ _ 2 ways
3 1 2 6 _ _ 2 ways
3 2 1 6 _ _ 2 ways
ways
If is the fifth number, then we cannot have the following:
_ _ _ _ 6 5 24 ways
1 5 _ _ 6 _ 6 ways
1 _ 5 _ 6 _ 6 ways
2 1 5 _ 6 _ 2 ways
1 _ _ 5 6 _ 6 ways
2 1 _ 5 6 _ 2 ways
2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 3 ways
ways
Grand Total :
Solution 3 (needs explanation)
The answer is .
Solution 4 (You won't get 458, 459, 460, 462, 465, 467, etc. with this method!!!)
First let us look at the General Case of this kind of Permutation: Consider this kind of Permutation of set for arbitrary
It is easy to count the total number of the permutation () of :
For every , we can divide into two subsets:
Define permutation as the permutation satisfy the condition of this problem. Then according to the condition of this problem, for each , is not a permutation of set . For each , mark the number of permutation of set as , where , mark the number of permutation for set as ; then, according to the condition of this problem, the permutation for is unrestricted, so the number of the unrestricted permutation of is . As a result, for each , the total number of permutation is
Notice that according to the condition of this problem, if you sum all up, you will get the total number of permutation of , that is,
Put , we will have
So the total number of permutations satisify this problem is .
~Solution by (Frank FYC)
2018 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 12 | |
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