2018 AIME II Problems/Problem 11
Contents
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
or
, 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 :
Solution 3 (needs explanation)
The answer is .
Solution 4 (You won't get any other answer with this method!!!)
First let us look at the General Case of this kind of Permutation: Consider this kind of Permutation of set for arbitrary
It is easy to count the total number of the permutation () of
:
For every , we can divide
into two parts:
Define permutation as the permutation satisfy the condition of this problem. Then according to the condition of this problem, for each
,
is not a permutation of set
. For each
, mark the number of permutation
of set
as
, where
, mark the number of permutation
for set
as
; then, according to the condition of this problem, the permutation for
is unrestricted, so the number of the unrestricted permutation of
is
. As a result, for each
, the total number of permutation
is
Notice that according to the condition of this problem, if you sum up all up, you will get the total number of permutation of
, that is,
Put , we will have
~Solution by (Frank FYC)
2018 AIME II (Problems • Answer Key • Resources) | ||
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