Difference between revisions of "2018 AIME II Problems/Problem 11"
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2 3 _ 6 _ _ <math>\implies</math> 4 ways | 2 3 _ 6 _ _ <math>\implies</math> 4 ways | ||
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2 4 _ 6 _ _ <math>\implies</math> 6 ways | 2 4 _ 6 _ _ <math>\implies</math> 6 ways | ||
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− | <math>24 + 18 | + | <math>24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71</math> ways |
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If <math>6</math> is the fourth number, then we cannot have the following: | If <math>6</math> is the fourth number, then we cannot have the following: | ||
+ | |||
1 _ _ 6 _ _ <math>\implies</math> 24 ways | 1 _ _ 6 _ _ <math>\implies</math> 24 ways | ||
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Grand Total : <math>120 + 96 + 90 + 84 + 71 = \boxed{461}</math> | Grand Total : <math>120 + 96 + 90 + 84 + 71 = \boxed{461}</math> | ||
+ | |||
+ | ==Solution 3 (Recursion)== | ||
+ | |||
+ | Note the condition in the problem is equivalent to the following condition: for each <math>k</math> with <math>1 \le k \le 5</math>, the first <math>k</math> terms is not a permutation <math>(1, 2, \ldots, k)</math> (since it would mean there must be some integer <math>x</math> in the first <math>k</math> terms such that <math>x \not \in \{1, 2, \ldots, k\} \implies x > k</math>). Then, let <math>a_n</math> denote the number of permutations of <math>(1, 2, \ldots, n)</math> satisfying the condition in the problem. We use complementary counting to find <math>a_n</math>. Notice that in order to not satisfy the condition in the problem, there are <math>n-1</math> cases: the first <math>1 \le k \le n-1</math> (we don't include <math>k = n</math> since the condition in the problem only holds up to <math>n-1</math>) terms are a permutation of <math>(1, 2, \ldots, k)</math>, and for all <math>k+1 \le i \le n-1</math>, the condition in the problem still holds. Then, for each of these cases, there are <math>k!</math> ways to arrange the first <math>k</math> terms, and then <math>a_{n-k}</math> ways to arrange the <math>k + 1</math> to <math>n</math> terms (assume by contradiction that terms from <math>k+1</math> to <math>i</math> is a permutation of <math>(k+1, k+2, \ldots, i)</math>. Then, since the first <math>k</math> terms are already a permutation of <math>(1, 2, \ldots, k)</math>, the first <math>i</math> terms form a permutation of <math>(1, 2, \ldots, i)</math>. This contradicts our assumption that for all <math>k+1 \le i \le n-1</math>, the condition still holds. Thus, we can only include the <math>a_{n-k}</math> permutations of these terms). Now, summing the cases up and subtracting from <math>n!</math>, we have: <math>a_n = n! - \sum_{k=1}^{n-1} a_{n-k} k!</math>. From this recursion, we derive that | ||
+ | <math>a_1 = 1</math>, <math>a_2 = 1</math>, <math>a_3 = 3</math>, <math>a_4 = 13</math>, <math>a_5 = 71</math>, and finally <math>a_6 = \boxed{461}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez CrazyVideoGamez] | ||
+ | |||
+ | ~ <math>BladeRunnerAUG</math> (Frank FYC) | ||
+ | |||
+ | ==Solution 4 (PIE)== | ||
+ | Let <math>A_i</math> be the set of permutations such that there is no number greater than <math>i</math> in the first <math>i</math> places. Note that <math>\bigcap^{k}_{i=0}{A_{b_i}}=\prod^k_{i=1}{(b_i-b_{i-1})!}</math> for all <math>1\le b_0 < b_1\cdots < b_k \le 5</math> and that the set of restricted permutations is <math>A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5</math>. | ||
+ | |||
+ | We will compute the cardinality of this set with PIE. | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | &|A_1| + |A_2| + |A_3| + |A_4| + |A_5|\\ = &120 + 48 + 36 + 48 + 120 = 372\\ \\ | ||
+ | &|A_1 \cap A_2| + |A_1 \cap A_3| + |A_1 \cap A_4| + |A_1 \cap A_5| + |A_2 \cap A_3|\\ + &|A_2 \cap A_4| + |A_2 \cap A_5| + |A_3 \cap A_4| + |A_3 \cap A_5| + |A_4 \cap A_5|\\=&24+12+12+24+12+8+12+12+12+24=152\\ \\ | ||
+ | &|A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_2 \cap A_5| + |A_1 \cap A_3 \cap A_4| + |A_1 \cap A_3 \cap A_5|\\ +& |A_1 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_5| + |A_2 \cap A_4 \cap A_5| + |A_3 \cap A_4 \cap A_5|\\=&6 + 4 + 6 + 4 + 4 + 6 + 4 + 4 + 4 + 6 = 48\\ \\ | ||
+ | &|A_1 \cap A_2 \cap A_3 \cap A_4| + |A_1 \cap A_2 \cap A_3 \cap A_5| + |A_1 \cap A_2 \cap A_4 \cap A_5| + |A_1 \cap A_3 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4 \cap A_5|\\=&2 + 2 + 2 + 2 + 2 = 10\\ \\ | ||
+ | &|A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = 1 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | 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> | ||
+ | ~sigma | ||
+ | |||
+ | ==Solution 6 (Complementary)== | ||
+ | We can also solve this problem by counting the number of permutations that do NOT satisfy the given conditions; namely, these permutations must satisfy the condition that none of the first <math>k</math> terms is greater than <math>k</math> for <math>1\leq</math> <math>k</math> <math>\leq</math> <math>5</math>. We can further simplify this method by approaching it through casework on the first <math>k</math> terms. | ||
+ | |||
+ | Case 1: None of the first one terms is greater than 1 | ||
+ | |||
+ | The first term must obviously be one. Since there are five spaces left for numbers, there are a total of <math>5!=120</math> permutations for this case. | ||
+ | |||
+ | Case 2: None of the first two terms is greater than 2 | ||
+ | |||
+ | The first two terms must be 1 and 2 in some order. However, we already counted all cases starting with 1 in the previous case, so all of the permutations in this case must begin with <math>12\cdots</math>. Since there are four spaces left, there are a total of <math>4!=24</math> permutations for this case. | ||
+ | |||
+ | Case 3: None of the first three terms is greater than 3 | ||
+ | |||
+ | The first three terms must be 1, 2, and 3 in some order. However, the cases beginning with 1__ and 21_ have already been accounted for. There are now <math>3!-3 = 3</math> ways to order the first three numbers of the permutation, and <math>3!</math> ways to order the last three numbers, for a total of <math>3\times6 = 18</math> permutations. | ||
+ | |||
+ | Case 4: None of the first four terms is greater than 4 | ||
+ | |||
+ | You can see where the pattern is going - the first four terms must be 1, 2, 3, and 4 in some order. All cases starting with 1 (there are <math>3!=6</math>), the cases starting with 21 (there are <math>2!=2</math>), and the 3 cases from case 3 (there are <math>3\times 1! = 3</math>) have been accounted for, so there are a total of <math>(4!-6-2-3)2!=26</math> permutations for this case. | ||
+ | |||
+ | Case 5: None of the first five terms is greater than 5 | ||
+ | |||
+ | This is perhaps the hardest case to work with, simply because there are so many subcases, so keeping track is crucial here. Obviously, the first five terms must be 1, 2, 3, 4, and 5, meaning there are 120 ways to order them. Now, we count the permutations we have already counted in previous cases. <math>4!</math> start with 1, <math>3!</math> start with 2, <math>3\times2!=6</math> start with 3, and <math>13\times1!=13</math> start with 4. Subtracting, we get a total of <math>120-24-6-6-13=71</math> permutations. | ||
+ | |||
+ | Now, we subtract the total number of permutations from our cases from the total number of permutations, which is <math>6!</math> : | ||
+ | <math>720 - 120 - 24 - 18 - 26 - 71 = \boxed{461}</math>. | ||
+ | |||
+ | ~TGSN/curiousmind_888 | ||
+ | |||
{{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}} |
Latest revision as of 13:13, 24 January 2024
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 (Recursion)
Note the condition in the problem is equivalent to the following condition: for each with
, the first
terms is not a permutation
(since it would mean there must be some integer
in the first
terms such that
). Then, let
denote the number of permutations of
satisfying the condition in the problem. We use complementary counting to find
. Notice that in order to not satisfy the condition in the problem, there are
cases: the first
(we don't include
since the condition in the problem only holds up to
) terms are a permutation of
, and for all
, the condition in the problem still holds. Then, for each of these cases, there are
ways to arrange the first
terms, and then
ways to arrange the
to
terms (assume by contradiction that terms from
to
is a permutation of
. Then, since the first
terms are already a permutation of
, the first
terms form a permutation of
. This contradicts our assumption that for all
, the condition still holds. Thus, we can only include the
permutations of these terms). Now, summing the cases up and subtracting from
, we have:
. From this recursion, we derive that
,
,
,
,
, and finally
.
~ (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
~sigma
Solution 6 (Complementary)
We can also solve this problem by counting the number of permutations that do NOT satisfy the given conditions; namely, these permutations must satisfy the condition that none of the first terms is greater than
for
. We can further simplify this method by approaching it through casework on the first
terms.
Case 1: None of the first one terms is greater than 1
The first term must obviously be one. Since there are five spaces left for numbers, there are a total of permutations for this case.
Case 2: None of the first two terms is greater than 2
The first two terms must be 1 and 2 in some order. However, we already counted all cases starting with 1 in the previous case, so all of the permutations in this case must begin with . Since there are four spaces left, there are a total of
permutations for this case.
Case 3: None of the first three terms is greater than 3
The first three terms must be 1, 2, and 3 in some order. However, the cases beginning with 1__ and 21_ have already been accounted for. There are now ways to order the first three numbers of the permutation, and
ways to order the last three numbers, for a total of
permutations.
Case 4: None of the first four terms is greater than 4
You can see where the pattern is going - the first four terms must be 1, 2, 3, and 4 in some order. All cases starting with 1 (there are ), the cases starting with 21 (there are
), and the 3 cases from case 3 (there are
) have been accounted for, so there are a total of
permutations for this case.
Case 5: None of the first five terms is greater than 5
This is perhaps the hardest case to work with, simply because there are so many subcases, so keeping track is crucial here. Obviously, the first five terms must be 1, 2, 3, 4, and 5, meaning there are 120 ways to order them. Now, we count the permutations we have already counted in previous cases. start with 1,
start with 2,
start with 3, and
start with 4. Subtracting, we get a total of
permutations.
Now, we subtract the total number of permutations from our cases from the total number of permutations, which is :
.
~TGSN/curiousmind_888
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 | ||
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