Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2018 AIME II Problems/Problem 11"

(→‎Solution 4 (You won't get any other answer with this method!!!))
(→‎Solution 4 (You won't get any other answer with this method!!!))
Line 126: Line 126:
 
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!!!)==
+
==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 <cmath>S=\{1,2,...,n\}</cmath> for arbitrary <math>n \in N</math>
 
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>
Line 132: Line 132:
 
It is easy to count the total number of the permutation (<math>N</math>) of <math>S</math>: <cmath>N=n!</cmath>
 
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>
+
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-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>
 
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>

Revision as of 10:14, 6 October 2018

Problem

Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$, at least one of the first $k$ terms of the permutation is greater than $k$.

Solution 1

If the first number is $6$, then there are no restrictions. There are $5!$, or $120$ ways to place the other $5$ numbers.


If the first number is $5$, $6$ can go in four places, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.


If the first number is $4$, ....

4 6 _ _ _ _ $\implies$ 24 ways

4 _ 6 _ _ _ $\implies$ 24 ways

4 _ _ 6 _ _ $\implies$ 24 ways

4 _ _ _ 6 _ $\implies$ 5 must go between $4$ and $6$, so there are $3 \cdot 3! = 18$ ways.

$24 + 24 + 24 + 18 = 90$ ways if 4 is first.


If the first number is $3$, ....

3 6 _ _ _ _ $\implies$ 24 ways

3 _ 6 _ _ _ $\implies$ 24 ways

3 1 _ 6 _ _ $\implies$ 4 ways

3 2 _ 6 _ _ $\implies$ 4 ways

3 4 _ 6 _ _ $\implies$ 6 ways

3 5 _ 6 _ _ $\implies$ 6 ways

3 5 _ _ 6 _ $\implies$ 6 ways

3 _ 5 _ 6 _ $\implies$ 6 ways

3 _ _ 5 6 _ $\implies$ 4 ways

$24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84$ ways


If the first number is $2$, ....

2 6 _ _ _ _ $\implies$ 24 ways

2 _ 6 _ _ _ $\implies$ 18 ways

2 3 _ 6 _ _ $\implies$ 4 ways

2 4 _ 6 _ _ $\implies$ 4 ways

2 4 _ 6 _ _ $\implies$ 6 ways

2 5 _ 6 _ _ $\implies$ 6 ways

2 5 _ _ 6 _ $\implies$ 6 ways

2 _ 5 _ 6 _ $\implies$ 4 ways

2 4 _ 5 6 _ $\implies$ 2 ways

2 3 4 5 6 1 $\implies$ 1 way


$24 + 18 + 4 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71$ ways


Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

Solution 2

If $6$ is the first number, then there are no restrictions. There are $5!$, or $120$ ways to place the other $5$ numbers.


If $6$ is the second number, then the first number can be $2, 3, 4,$ or $5$, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.


If $6$ is the third number, then we cannot have the following:

1 _ 6 _ _ _ $\implies$ 24 ways

2 1 6 _ _ _ $\implies$ 6 ways

$120 - 24 - 6 = 90$ ways

If $6$ is the fourth number, then we cannot have the following: 1 _ _ 6 _ _ $\implies$ 24 ways

2 1 _ 6 _ _ $\implies$ 6 ways

2 3 1 6 _ _ $\implies$ 2 ways

3 1 2 6 _ _ $\implies$ 2 ways

3 2 1 6 _ _ $\implies$ 2 ways

$120 - 24 - 6 - 2 - 2 - 2 = 84$ ways

If $6$ is the fifth number, then we cannot have the following:

_ _ _ _ 6 5 $\implies$ 24 ways

1 5 _ _ 6 _ $\implies$ 6 ways

1 _ 5 _ 6 _ $\implies$ 6 ways

2 1 5 _ 6 _ $\implies$ 2 ways

1 _ _ 5 6 _ $\implies$ 6 ways

2 1 _ 5 6 _ $\implies$ 2 ways

2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 $\implies$ 3 ways

$120 - 24 - 6 - 6 - 2 - 6 - 2 - 3 = 71$ ways

Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

Solution 3 (needs explanation)

The answer is $\frac{6!}{2} + 5! - 4! + 3! - 2! + 1! = \boxed{\boxed{461}}$.

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 \[S=\{1,2,...,n\}\] for arbitrary $n \in N$

It is easy to count the total number of the permutation ($N$) of $S$: \[N=n!\]

For every $i \in S$, we can divide $S$ into two subsets: \[S_{1\to i}=\{1,2,...i\}; S_{i+1\to n}=\{i+1,i+2,...,n\}\]

Define permutation $P$ as the permutation satisfy the condition of this problem. Then according to the condition of this problem, for each $i\in \{1,2,...,n-1\}$, $P$ is not a permutation of set $S_{1\to i}$. For each $i\in \{1,2,...,n-1\}$, mark the number of permutation $P$ of set $S$ as $P_{k}$, where $k=i$, mark the number of permutation $P$ for set $S_{i+1\to n}$ as $P_{i}$; then, according to the condition of this problem, the permutation for $S_{i+1\to n}$ is unrestricted, so the number of the unrestricted permutation of $S_{i+1\to n}$ is $(n-i)!$. As a result, for each $i\in \{1,2,...,n-1\}$, the total number of permutation $P$ is \[P_{k}=P_{i}(n-i)!\]

Notice that according to the condition of this problem, if you sum up all $P_{k}$ up, you will get the total number of permutation of $S$, that is, \[N=\sum^{n}_{k=1}{P_{k}}=\sum^{n}_{i=1}{P_{i}(n-i)!}=n!\]

Put $n=1,2,3,...,6$, we will have

\[P_{1}=1\]

\[P_{2}=1\]

\[P_{3}=3\]

\[P_{4}=13\]

\[P_{5}=71\]

\[P_{6}=\boxed{461}\]

~Solution by $BladeRunnerAUG$ (Frank FYC)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_11&oldid=98028"