Difference between revisions of "2018 AIME II Problems/Problem 11"
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The answer is <math>\frac{6!}{2} + 5! - 4! + 3! - 2! + 1! = \boxed{\boxed{461}}</math>. | The answer is <math>\frac{6!}{2} + 5! - 4! + 3! - 2! + 1! = \boxed{\boxed{461}}</math>. | ||
+ | ==Solution 4 (You won't get any other answer with this method!!!)== | ||
+ | |||
+ | First let us look at the General Case of this kind of Permutation: Consider this kind of Permutation of set <cmath>S=\{1,2,...,n\}</cmath> for arbitrary <math>n \in N</math> | ||
+ | |||
+ | It is easy to count the total number of the permutation (<math>N</math>) of <math>S</math>: <cmath>N=n!</cmath> | ||
+ | |||
+ | For every <math>i \in S</math>, we can divide <math>S</math> into two parts: <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-1\}</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-1\}</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 up 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> | ||
+ | |||
+ | Put <math>n=1,2,3,...,6</math>, we will have | ||
+ | |||
+ | <cmath>P_{1}=1</cmath> | ||
+ | |||
+ | <cmath>P_{2}=1</cmath> | ||
+ | |||
+ | <cmath>P_{3}=3</cmath> | ||
+ | |||
+ | <cmath>P_{4}=13</cmath> | ||
+ | |||
+ | <cmath>P_{5}=71</cmath> | ||
+ | |||
+ | <cmath>P_{6}=\boxed{461}</cmath> | ||
{{AIME box|year=2018|n=II|num-b=10|num-a=12}} | {{AIME box|year=2018|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:11, 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 any other answer 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 parts:
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 up all up, you will get the total number of permutation of , that is,
Put , we will have
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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