Difference between revisions of "2018 AIME II Problems/Problem 11"
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To finish, <math>720 - 372 + 152 - 48 + 10 - 1 = \boxed{461}</math> | To finish, <math>720 - 372 + 152 - 48 + 10 - 1 = \boxed{461}</math> | ||
+ | ==Solution 5 (Recursion)== | ||
+ | Define the function <math>f(p,q)</math> as the amount of permutations with maximum digit <math>q</math> and string length <math>p</math> that satisfy the condition within bounds. For example, <math>f(4,5)</math> would be the amount of ways to make a string with length <math>4</math> with the highest digit being <math>5</math>. We wish to obtain <math>f(6,6)=f(5,6)</math>. | ||
+ | |||
+ | To generate recursion, consider how we would get to <math>f(p,q)</math> from <math>f(p-1,a)</math> for all <math>a</math> such that <math>p\le{a}\le6</math>. We could either jump from the old maximum <math>a</math> to the new <math>q</math> by concatenating the old string and the new digit <math>q</math>, or one could retain the maximum, in which case <math>a=q</math>. To retain the maximum, one would have to pick a new available digit not exceeding <math>q</math>. | ||
+ | |||
+ | In the first case, there is only one way to pick the new digit, namely picking <math>q</math>. For the second case, there are <math>q-p+1</math> digits left to choose, because there are <math>q</math> digits between 1 and <math>q</math> total and there are <math>p-1</math> digits already chosen below or equal to <math>q</math>. Thus, <math>f(p,q)=[\sum^{q-1}_{n=p}f(p-1,n)] + (q-p+1)f(p-1,q)</math>. Now that we have the recursive function, we can start evaluating the values of <math>f(p,q)</math> until we get to <math>f(6,6)=f(5,6)</math>. | ||
+ | |||
+ | <cmath>f(2,3)=3, f(2,4)=5, f(2,5)=7, f(2,6)=9</cmath> | ||
+ | <cmath>f(3,4)=13, f(3,5)=29, f(3,6)=51</cmath> | ||
+ | <cmath>f(4,5)=71, f(4,6)=195</cmath> | ||
+ | <cmath>f(5,6)=461</cmath> | ||
+ | Our requested answer is thus <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 00:02, 15 May 2022
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 _ _ 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 (General Case)
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 subsets:
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 all
up, you will get the total number of permutation of
, that is,
Put
, we will have
So the total number of permutations satisify this problem is
.
~Solution by (Frank FYC)
Solution 4 (PIE)
Let be the set of permutations such that there is no number greater than
in the first
places. Note that
for all
and that the set of restricted permutations is
.
We will compute the cardinality of this set with PIE.
To finish,
Solution 5 (Recursion)
Define the function as the amount of permutations with maximum digit
and string length
that satisfy the condition within bounds. For example,
would be the amount of ways to make a string with length
with the highest digit being
. We wish to obtain
.
To generate recursion, consider how we would get to from
for all
such that
. We could either jump from the old maximum
to the new
by concatenating the old string and the new digit
, or one could retain the maximum, in which case
. To retain the maximum, one would have to pick a new available digit not exceeding
.
In the first case, there is only one way to pick the new digit, namely picking . For the second case, there are
digits left to choose, because there are
digits between 1 and
total and there are
digits already chosen below or equal to
. Thus,
. Now that we have the recursive function, we can start evaluating the values of
until we get to
.
Our requested answer is thus
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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