Difference between revisions of "2018 AIME II Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | 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 <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 5, 6 can go in four places, and there are 4! ways to place the other 4 numbers. 4 | + | If the first number is <math>5</math>, <math>6</math> can go in four places, and there are <math>4!</math> ways to place the other <math>4</math> numbers. <math>4 \cdot 4! = 96</math> ways. |
− | If the first number is 4, .... | + | If the first number is <math>4</math>, .... |
− | 4 6 _ _ _ _ | + | 4 6 _ _ _ _ <math>\implies</math> 24 ways |
− | 4 _ 6 _ _ _ | + | 4 _ 6 _ _ _ <math>\implies</math> 24 ways |
− | 4 _ _ 6 _ _ | + | 4 _ _ 6 _ _ <math>\implies</math> 24 ways |
− | 4 _ _ _ 6 _ | + | 4 _ _ _ 6 _ <math>\implies</math> 5 must go between <math>4</math> and <math>6</math>, so there are <math>3 \cdot 3! = 18</math> ways. |
24 + 24 + 24 + 18 = 90 ways if 4 is first. | 24 + 24 + 24 + 18 = 90 ways if 4 is first. | ||
− | If the first number is 3, .... | + | If the first number is <math>3</math>, .... |
− | 3 6 _ _ _ _ | + | 3 6 _ _ _ _ <math>\implies</math> 24 ways |
− | 3 _ 6 _ _ _ | + | 3 _ 6 _ _ _ <math>\implies</math> 24 ways |
− | 3 1 _ 6 _ _ | + | 3 1 _ 6 _ _ <math>\implies</math> 4 ways |
− | 3 2 _ 6 _ _ | + | 3 2 _ 6 _ _ <math>\implies</math> 4 ways |
− | 3 4 _ 6 _ _ | + | 3 4 _ 6 _ _ <math>\implies</math> 6 ways |
− | 3 5 _ 6 _ _ | + | 3 5 _ 6 _ _ <math>\implies</math> 6 ways |
− | 3 5 _ _ 6 _ | + | 3 5 _ _ 6 _ <math>\implies</math> 6 ways |
− | 3 _ 5 _ 6 _ | + | 3 _ 5 _ 6 _ <math>\implies</math> 6 ways |
− | 3 _ _ 5 6 _ | + | 3 _ _ 5 6 _ <math>\implies</math> 4 ways |
− | 24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84 ways | + | <math>24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84</math> ways |
− | If the first number is 2, .... | + | If the first number is <math>2</math>, .... |
− | 2 6 _ _ _ _ | + | 2 6 _ _ _ _ <math>\implies</math> 24 ways |
− | 2 _ 6 _ _ _ | + | 2 _ 6 _ _ _ <math>\implies</math> 18 ways |
− | 2 3 _ 6 _ _ | + | 2 3 _ 6 _ _ <math>\implies</math> 4 ways |
− | 2 4 _ 6 _ _ | + | 2 4 _ 6 _ _ <math>\implies</math> 4 ways |
− | 2 4 _ 6 _ _ | + | 2 4 _ 6 _ _ <math>\implies</math> 6 ways |
− | 2 5 _ 6 _ _ | + | 2 5 _ 6 _ _ <math>\implies</math> 6 ways |
− | 2 5 _ _ 6 _ | + | 2 5 _ _ 6 _ <math>\implies</math> 6 ways |
− | 2 _ 5 _ 6 _ | + | 2 _ 5 _ 6 _ <math>\implies</math> 4 ways |
− | 2 4 _ 5 6 _ | + | 2 4 _ 5 6 _ <math>\implies</math> 2 ways |
− | 2 3 4 5 6 1 | + | 2 3 4 5 6 1 <math>\implies</math> 1 way |
− | 24 + 18 + 4 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71 ways | + | <math>24 + 18 + 4 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71</math> ways |
− | Grand Total : 120 + 96 + 90 + 84 + 71 = <math>\boxed{461}</math> | + | Grand Total : <math>120 + 96 + 90 + 84 + 71 = </math><math>\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}} |
Revision as of 20:24, 26 March 2018
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
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.
24 + 24 + 24 + 18 = 90 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 :
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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