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 458, 459, 460, 462, 465, 467, etc. with this method!!!))
m (→‎Solution 1)
Line 54: Line 54:
  
 
2 3 _ 6 _ _ <math>\implies</math> 4 ways
 
2 3 _ 6 _ _ <math>\implies</math> 4 ways
−
 
−
2 4 _ 6 _ _ <math>\implies</math> 4 ways
 
  
 
2 4 _ 6 _ _ <math>\implies</math> 6 ways
 
2 4 _ 6 _ _ <math>\implies</math> 6 ways
Line 70: Line 68:
  
  
−
<math>24 + 18 + 4 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71</math> ways
+
<math>24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71</math> ways
  
  

Revision as of 00:38, 26 December 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 _ _ $\implies$ 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


$24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71$ 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 (General Case, and 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 $i\in \{1,2,...,n\}$, 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 $i\in \{1,2,...,n\}$, 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 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 \[P_{1}=1\] \[P_{2}=1\] \[P_{3}=3\] \[P_{4}=13\] \[P_{5}=71\] \[P_{6}=461\] So the total number of permutations satisify this problem is $P_{6}=\boxed{461}$.

~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=99764"