2018 AIME II Problems/Problem 11: Difference between revisions
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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>. | ||
==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 5, 6 can go in four places, and there are 4! ways to place the other 4 numbers. 4 * 4! = 96 ways. | |||
If the first number is 4, .... | |||
4 6 _ _ _ _ -> 24 ways | |||
4 _ 6 _ _ _ -> 24 ways | |||
4 _ _ 6 _ _ -> 24 ways | |||
4 _ _ _ 6 _ -> 5 must go between 4 and 6, so there are 3 * 3! = 18 ways. | |||
24 + 24 + 24 + 18 = 90 ways if 4 is first. | |||
If the first number is 3, .... | |||
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 | |||
24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84 ways | |||
If the first number is 2, .... | |||
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 | |||
24 + 18 + 4 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71 ways | |||
Grand Total : 120 + 96 + 90 + 84 + 71 = <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 12:15, 24 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 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 * 4! = 96 ways.
If the first number is 4, ....
4 6 _ _ _ _ -> 24 ways
4 _ 6 _ _ _ -> 24 ways
4 _ _ 6 _ _ -> 24 ways
4 _ _ _ 6 _ -> 5 must go between 4 and 6, so there are 3 * 3! = 18 ways.
24 + 24 + 24 + 18 = 90 ways if 4 is first.
If the first number is 3, ....
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
24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84 ways
If the first number is 2, ....
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
24 + 18 + 4 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71 ways
Grand Total : 120 + 96 + 90 + 84 + 71 = 
| 2018 AIME II (Problems • Answer Key • Resources) | ||
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Followed by Problem 12 | |
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