Difference between revisions of "2018 AIME II Problems/Problem 11"
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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. | 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. | + | <math>24 + 24 + 24 + 18 = 90</math> ways if 4 is first. |
Revision as of 09:38, 8 April 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.
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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