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

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Find the number of permutations of <math>1, 2, 3, 4, 5, 6</math> such that for each <math>k</math> with <math>1</math> <math>\leq</math> <math>k</math> <math>\leq</math> <math>5</math>, at least one of the first <math>k</math> terms of the permutation is greater than <math>k</math>.
 
Find the number of permutations of <math>1, 2, 3, 4, 5, 6</math> such that for each <math>k</math> with <math>1</math> <math>\leq</math> <math>k</math> <math>\leq</math> <math>5</math>, at least one of the first <math>k</math> terms of the permutation is greater than <math>k</math>.
  
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==Solution==
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==Solution 1==
  
 
If the first number is <math>6</math>, then there are no restrictions. There are <math>5!</math>, or <math>120</math> ways to place the other <math>5</math> numbers.
 
If the first number is <math>6</math>, then there are no restrictions. There are <math>5!</math>, or <math>120</math> ways to place the other <math>5</math> numbers.
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Grand Total : <math>120 + 96 + 90 + 84 + 71 = </math><math>\boxed{461}</math>
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Grand Total : <math>120 + 96 + 90 + 84 + 71 = \boxed{461}</math>
  
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{{AIME box|year=2018|n=II|num-b=10|num-a=12}}
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==Solution 2==
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{{MAA Notice}}
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If <math>6</math> is the first number, then there are no restrictions. There are <math>5!</math>, or <math>120</math> ways to place the other <math>5</math> numbers.
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 +
 
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If <math>6</math> is the second number, then the first number can be <math>2, 3, 4, or 5</math>, and there are <math>4!</math> ways to place the other <math>4</math> numbers. <math>4 \cdot 4! = 96</math> ways.
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If <math>6</math> is the third number, then we cannot have the following:
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1 _ 6 _ _ _ <math>\implies</math> 24 ways
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2 1 6 _ _ _ <math>\implies</math> 6 ways
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<math>120 - 24 - 6 = 90</math> ways
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If <math>6</math> is the fourth number, then we cannot have the following:
 +
1 _ _ 6 _ _ <math>\implies</math> 24 ways
 +
 
 +
2 1 _ 6 _ _ <math>\implies</math> 6 ways
 +
 
 +
2 3 1 6 _ _ <math>\implies</math> 2 ways
 +
 
 +
3 1 2 6 _ _ <math>\implies</math> 2 ways
 +
 
 +
3 2 1 6 _ _ <math>\implies</math> 2 ways
 +
 
 +
<math>120 - 24 - 6 - 2 - 2 - 2 = 84</math> ways
 +
 
 +
If <math>6</math> is the fifth number, then we cannot have the following:
 +
 
 +
_ _ _ _ 6 5 <math>\implies</math> 24 ways
 +
 
 +
1 5 _ _ 6 _ <math>\implies</math> 6 ways
 +
 
 +
1 _ 5 _ 6 _ <math>\implies</math> 6 ways
 +
 
 +
2 1 5 _ 6 _ <math>\implies</math> 2 ways
 +
 
 +
1 _ _ 5 6 _ <math>\implies</math> 6 ways
 +
 
 +
2 1 _ 5 6 _ <math>\implies</math> 2 ways
 +
 
 +
2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 <math>\implies</math> 3 ways
 +
 
 +
<math>120 - 24 - 6 - 6 - 2 - 6 - 2 - 3 = 71</math> ways
 +
 
 +
Grand Total : <math>120 + 96 + 90 + 84 + 71 = \boxed{461}</math>

Revision as of 15:06, 14 April 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 : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

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