Difference between revisions of "2018 AIME II Problems/Problem 11"
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2 3 _ 6 _ _ <math>\implies</math> 4 ways | 2 3 _ 6 _ _ <math>\implies</math> 4 ways | ||
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2 4 _ 6 _ _ <math>\implies</math> 6 ways | 2 4 _ 6 _ _ <math>\implies</math> 6 ways | ||
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− | <math>24 + 18 | + | <math>24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71</math> ways |
Line 91: | Line 89: | ||
If <math>6</math> is the fourth number, then we cannot have the following: | If <math>6</math> is the fourth number, then we cannot have the following: | ||
+ | |||
1 _ _ 6 _ _ <math>\implies</math> 24 ways | 1 _ _ 6 _ _ <math>\implies</math> 24 ways | ||
Line 123: | Line 122: | ||
Grand Total : <math>120 + 96 + 90 + 84 + 71 = \boxed{461}</math> | Grand Total : <math>120 + 96 + 90 + 84 + 71 = \boxed{461}</math> | ||
− | ==Solution 3 ( | + | ==Solution 3 (General Case)== |
− | + | ||
+ | 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 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\}</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> 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}=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 4 (PIE)== | ||
+ | Let <math>A_i</math> be the set of permutations such that there is no number greater than <math>i</math> in the first <math>i</math> places. Note that <math>\bigcap^{k}_{i=0}{A_{b_i}}=\prod^k_{i=1}{(b_i-b_{i-1})!}</math> for all <math>1\le b_0 < b_1\cdots < b_k \le 5</math> and that the set of restricted permutations is <math>A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5</math>. | ||
+ | |||
+ | We will compute the cardinality of this set with PIE. | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | &|A_1| + |A_2| + |A_3| + |A_4| + |A_5|\\ = &120 + 48 + 36 + 48 + 120 = 372\\ \\ | ||
+ | &|A_1 \cap A_2| + |A_1 \cap A_3| + |A_1 \cap A_4| + |A_1 \cap A_5| + |A_2 \cap A_3|\\ + &|A_2 \cap A_4| + |A_2 \cap A_5| + |A_3 \cap A_4| + |A_3 \cap A_5| + |A_4 \cap A_5|\\=&24+12+12+24+12+8+12+12+12+24=152\\ \\ | ||
+ | &|A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_2 \cap A_5| + |A_1 \cap A_3 \cap A_4| + |A_1 \cap A_3 \cap A_5|\\ +& |A_1 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_5| + |A_2 \cap A_4 \cap A_5| + |A_3 \cap A_4 \cap A_5|\\=&6 + 4 + 6 + 4 + 4 + 6 + 4 + 4 + 4 + 6 = 48\\ \\ | ||
+ | &|A_1 \cap A_2 \cap A_3 \cap A_4| + |A_1 \cap A_2 \cap A_3 \cap A_5| + |A_1 \cap A_2 \cap A_4 \cap A_5| + |A_1 \cap A_3 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4 \cap A_5|\\=&2 + 2 + 2 + 2 + 2 = 10\\ \\ | ||
+ | &|A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = 1 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | To finish, <math>720 - 372 + 152 - 48 + 10 - 1 = \boxed{461}</math> | ||
{{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}} |
Latest revision as of 18:22, 24 February 2021
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 _ _ 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 (General Case)
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)
Solution 4 (PIE)
Let be the set of permutations such that there is no number greater than in the first places. Note that for all and that the set of restricted permutations is .
We will compute the cardinality of this set with PIE. To finish,
2018 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 12 | |
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