# 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 1

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 :

## Solution 2

If is the first number, then there are no restrictions. There are , or ways to place the other numbers.

If is the second number, then the first number can be , and there are ways to place the other numbers. ways.

If is the third number, then we cannot have the following:

1 _ 6 _ _ _ 24 ways

2 1 6 _ _ _ 6 ways

ways

If is the fourth number, then we cannot have the following: 1 _ _ 6 _ _ 24 ways

2 1 _ 6 _ _ 6 ways

2 3 1 6 _ _ 2 ways

3 1 2 6 _ _ 2 ways

3 2 1 6 _ _ 2 ways

ways

If is the fifth number, then we cannot have the following:

_ _ _ _ 6 5 24 ways

1 5 _ _ 6 _ 6 ways

1 _ 5 _ 6 _ 6 ways

2 1 5 _ 6 _ 2 ways

1 _ _ 5 6 _ 6 ways

2 1 _ 5 6 _ 2 ways

2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 3 ways

ways

Grand Total :