2018 AIME II Problems/Problem 11
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) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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