AIME 2018 II · 第 11 题

Solution 1

If the first number is $6$, then there are no restrictions. There are $5!$, or $120$ ways to place the other $5$ numbers.

If the first number is $5$, $6$ can go in four places, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.

If the first number is $4$, ....

4 6 _ _ _ _ $\implies$ 24 ways

4 _ 6 _ _ _ $\implies$ 24 ways

4 _ _ 6 _ _ $\implies$ 24 ways

4 _ _ _ 6 _ $\implies$ 5 must go between $4$ and $6$, so there are $3 \cdot 3! = 18$ ways.

$24 + 24 + 24 + 18 = 90$ ways if 4 is first.

If the first number is $3$, ....

3 6 _ _ _ _ $\implies$ 24 ways

3 _ 6 _ _ _ $\implies$ 24 ways

3 1 _ 6 _ _ $\implies$ 4 ways

3 2 _ 6 _ _ $\implies$ 4 ways

3 4 _ 6 _ _ $\implies$ 6 ways

3 5 _ 6 _ _ $\implies$ 6 ways

3 5 _ _ 6 _ $\implies$ 6 ways

3 _ 5 _ 6 _ $\implies$ 6 ways

3 _ _ 5 6 _ $\implies$ 4 ways

$24 + 24 + 4 + 4 + 6 + 6 + 6 + 6 + 4 = 84$ ways

If the first number is $2$, ....

2 6 _ _ _ _ $\implies$ 24 ways

2 _ 6 _ _ _ $\implies$ 18 ways

2 3 _ 6 _ _ $\implies$ 4 ways

2 4 _ 6 _ _ $\implies$ 6 ways

2 5 _ 6 _ _ $\implies$ 6 ways

2 5 _ _ 6 _ $\implies$ 6 ways

2 _ 5 _ 6 _ $\implies$ 4 ways

2 4 _ 5 6 _ $\implies$ 2 ways

2 3 4 5 6 1 $\implies$ 1 way

$24 + 18 + 4 + 6 + 6 + 6 + 4 + 2 + 1 = 71$ ways

Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

Solution 2

If $6$ is the first number, then there are no restrictions. There are $5!$, or $120$ ways to place the other $5$ numbers.

If $6$ is the second number, then the first number can be $2, 3, 4,$ or $5$, and there are $4!$ ways to place the other $4$ numbers. $4 \cdot 4! = 96$ ways.

If $6$ is the third number, then we cannot have the following:

1 _ 6 _ _ _ $\implies$ 24 ways

2 1 6 _ _ _ $\implies$ 6 ways

$120 - 24 - 6 = 90$ ways

If $6$ is the fourth number, then we cannot have the following:

1 _ _ 6 _ _ $\implies$ 24 ways

2 1 _ 6 _ _ $\implies$ 6 ways

2 3 1 6 _ _ $\implies$ 2 ways

3 1 2 6 _ _ $\implies$ 2 ways

3 2 1 6 _ _ $\implies$ 2 ways

$120 - 24 - 6 - 2 - 2 - 2 = 84$ ways

If $6$ is the fifth number, then we cannot have the following:

_ _ _ _ 6 5 $\implies$ 24 ways

1 5 _ _ 6 _ $\implies$ 6 ways

1 _ 5 _ 6 _ $\implies$ 6 ways

2 1 5 _ 6 _ $\implies$ 2 ways

1 _ _ 5 6 _ $\implies$ 6 ways

2 1 _ 5 6 _ $\implies$ 2 ways

2 3 1 5 6 4, 3 1 2 5 6 4, 3 2 1 5 6 4 $\implies$ 3 ways

$120 - 24 - 6 - 6 - 2 - 6 - 2 - 3 = 71$ ways

Grand Total : $120 + 96 + 90 + 84 + 71 = \boxed{461}$

Solution 3 (Recursion)

Note the condition in the problem is equivalent to the following condition: for each $k$ with $1 \le k \le 5$, the first $k$ terms is not a permutation $(1, 2, \ldots, k)$ (since it would mean there must be some integer $x$ in the first $k$ terms such that $x \not \in \{1, 2, \ldots, k\} \implies x > k$). Then, let $a_n$ denote the number of permutations of $(1, 2, \ldots, n)$ satisfying the condition in the problem. We use complementary counting to find $a_n$. Notice that in order to not satisfy the condition in the problem, there are $n-1$ cases: the first $1 \le k \le n-1$ (we don't include $k = n$ since the condition in the problem only holds up to $n-1$) terms are a permutation of $(1, 2, \ldots, k)$, and for all $k+1 \le i \le n-1$, the condition in the problem still holds. Then, for each of these cases, there are $k!$ ways to arrange the first $k$ terms, and then $a_{n-k}$ ways to arrange the $k + 1$ to $n$ terms (assume by contradiction that terms from $k+1$ to $i$ is a permutation of $(k+1, k+2, \ldots, i)$. Then, since the first $k$ terms are already a permutation of $(1, 2, \ldots, k)$, the first $i$ terms form a permutation of $(1, 2, \ldots, i)$. This contradicts our assumption that for all $k+1 \le i \le n-1$, the condition still holds. Thus, we can only include the $a_{n-k}$ permutations of these terms). Now, summing the cases up and subtracting from $n!$, we have: $a_n = n! - \sum_{k=1}^{n-1} a_{n-k} k!$. From this recursion, we derive that $a_1 = 1$, $a_2 = 1$, $a_3 = 3$, $a_4 = 13$, $a_5 = 71$, and finally $a_6 = \boxed{461}$.

~CrazyVideoGamez

~ $BladeRunnerAUG$ (Frank FYC)

Solution 4 (PIE)

Let $A_i$ be the set of permutations such that there is no number greater than $i$ in the first $i$ places. Note that $\bigcap^{k}_{i=0}{A_{b_i}}=\prod^k_{i=1}{(b_i-b_{i-1})!}$ for all $1\le b_0 < b_1\cdots < b_k \le 5$ and that the set of restricted permutations is $A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5$.

We will compute the cardinality of this set with PIE.

$$\begin{aligned} &|A_1| + |A_2| + |A_3| + |A_4| + |A_5|\\ = &120 + 48 + 36 + 48 + 120 = 372\\ \\ &|A_1 \cap A_2| + |A_1 \cap A_3| + |A_1 \cap A_4| + |A_1 \cap A_5| + |A_2 \cap A_3|\\ + &|A_2 \cap A_4| + |A_2 \cap A_5| + |A_3 \cap A_4| + |A_3 \cap A_5| + |A_4 \cap A_5|\\=&24+12+12+24+12+8+12+12+12+24=152\\ \\ &|A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_2 \cap A_5| + |A_1 \cap A_3 \cap A_4| + |A_1 \cap A_3 \cap A_5|\\ +& |A_1 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_5| + |A_2 \cap A_4 \cap A_5| + |A_3 \cap A_4 \cap A_5|\\=&6 + 4 + 6 + 4 + 4 + 6 + 4 + 4 + 4 + 6 = 48\\ \\ &|A_1 \cap A_2 \cap A_3 \cap A_4| + |A_1 \cap A_2 \cap A_3 \cap A_5| + |A_1 \cap A_2 \cap A_4 \cap A_5| + |A_1 \cap A_3 \cap A_4 \cap A_5| + |A_2 \cap A_3 \cap A_4 \cap A_5|\\=&2 + 2 + 2 + 2 + 2 = 10\\ \\ &|A_1 \cap A_2 \cap A_3 \cap A_4 \cap A_5| = 1 \end{aligned}$$ To finish, $720 - 372 + 152 - 48 + 10 - 1 = \boxed{461}$

Solution 5 (Recursion)

Define the function $f(p,q)$ as the amount of permutations with maximum digit $q$ and string length $p$ that satisfy the condition within bounds. For example, $f(4,5)$ would be the amount of ways to make a string with length $4$ with the highest digit being $5$. We wish to obtain $f(6,6)=f(5,6)$.

To generate recursion, consider how we would get to $f(p,q)$ from $f(p-1,a)$ for all $a$ such that $p\le{a}\le6$. We could either jump from the old maximum $a$ to the new $q$ by concatenating the old string and the new digit $q$, or one could retain the maximum, in which case $a=q$. To retain the maximum, one would have to pick a new available digit not exceeding $q$.

In the first case, there is only one way to pick the new digit, namely picking $q$. For the second case, there are $q-p+1$ digits left to choose, because there are $q$ digits between 1 and $q$ total and there are $p-1$ digits already chosen below or equal to $q$. Thus, $f(p,q)=[\sum^{q-1}_{n=p}f(p-1,n)] + (q-p+1)f(p-1,q)$. Now that we have the recursive function, we can start evaluating the values of $f(p,q)$ until we get to $f(6,6)=f(5,6)$.

$$f(2,3)=3, f(2,4)=5, f(2,5)=7, f(2,6)=9$$ $$f(3,4)=13, f(3,5)=29, f(3,6)=51$$ $$f(4,5)=71, f(4,6)=195$$ $$f(5,6)=461$$ Our requested answer is thus $\boxed{461}$ ~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 $k$ terms is greater than $k$ for $1\leq$ $k$ $\leq$ $5$. We can further simplify this method by approaching it through casework on the first $k$ 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 $5!=120$ 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 $21\cdots$. Since there are four spaces left, there are a total of $4!=24$ 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 $3!-3 = 3$ ways to order the first three numbers of the permutation, and $3!$ ways to order the last three numbers, for a total of $3\times6 = 18$ 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 $3!=6$), the cases starting with 21 (there are $2!=2$), and the 3 cases from case 3 (there are $3\times 1! = 3$) have been accounted for, so there are a total of $(4!-6-2-3)2!=26$ 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. $4!$ start with 1, $3!$ start with 2, $3\times2!=6$ start with 3, and $13\times1!=13$ start with 4. Subtracting, we get a total of $120-24-6-6-13=71$ permutations.

Now, we subtract the total number of permutations from our cases from the total number of permutations, which is $6!$ : $720 - 120 - 24 - 18 - 26 - 71 = \boxed{461}$.

~TGSN/curiousmind_888

Solution 7(Similar to Solutions 1 and 2 but with more depth)

We will perform cases based on the position of $6$. Notice that if $6$ is the first number, the positions of the rest of the numbers absolutely don't matter, so we have $5!$ cases here.

Now, if $6$ is the second number, all we need is the first number to be greater than $1$. Namely, the possibilities for the first number are just $2$ to $5$, for a total of $4$ numbers. The rest of the number positions don't matter, so we have $4!$ ways to arrange the rest of the numbers. Hence, for this case, we have $4 \cdot 4!$ cases.

Now, if $6$ is the third number, we can break it up into two cases. First, the first number is simply $2$. Then, as "at least one of the first $2$ terms is greater than $2$" we just need the second term to be greater than $2$, namely the possibilities are $3$ to $5$, a total of $3$ numbers. The arrangement of the remaining numbers doesn't matter, and hence that is just $3!$. The first part of this case yields $3 \cdot 3!$. Now, the second part is that the first term is any number from $3$ to $5$. Then, the rest of the numbers don't matter. Hence, we have $4!$ ways to arrange the rest of the numbers. So we have a total of $3! \cdot 3 + 4! \cdot 3$ ways for this case.

Now, if $6$ is the fourth number, things get a little bit more difficult. We will break it up into $4$ cases. First, we have $2$ as the first term. Now, if the second term is $3$, all we need for this to succeed is for the third term to be at least $4$, so only two possibilities $4$ and $5$. The rest of the number arrangements don't matter, so we have $2!$ ways for that. So the first part of this case is $2 \cdot 2!$. Now, if the second term is at least $4$, then we are done. Hence, we have our two possibilities $4$ and $5$ for the second term, and the remaining number arrangements don't matter. Hence, we have $2 \cdot 3!$. So if the first term if $2$, we have $2 \cdot 2! + 2 \cdot 3!$ cases. Now, if the first term is $3$, all we need to satisfy is that "at least one of the first $3$ terms is greater than $3$." Hence, we can break this up also into cases. First, we'll consider two blocks: one block of numbers before the $6$ and one block of numbers after the $6$. If $4$ is in the first block before $6$ and $5$ is in the second block, we have $2!$ ways to arrange the rest of the numbers. There are $2$ ways to place the $4$ in the first block and $2$ ways to place the $5$ in the second block for a total of $2 \cdot 2 = 4$. We can also switch the positions of $4$ and $5$, namely, placing $5$ in the first block and $4$ in the second block, as this won't affect the requirements. So, we have a total of $2! \cdot 2 \cdot 4$ ways for the first part. Now, the second part is when both $4$ and $5$ are in the first block. There are $2$ ways for both $4$ and $5$ to be in the first block and then $2!$ ways to arrange the rest of the numbers. So we have $2! \cdot 2$ for this case. Hence, if $3$ is the first term, we have $2! \cdot 2 \cdot 4 + 2! \cdot 2$ cases. Now, if $4$ is the first term, everything is satisfied, and we are left with the task of finding how many ways there are to arrange the rest of the numbers. This is simply $4!$. The same occurs if $5$ is the first term with $4!$ ways.

After all the casework, if $6$ is the fourth number, there is a total of $2 \cdot 2! + 2 \cdot 3! + 2! \cdot 2 \cdot 4 + 2! \cdot 2 + 4! + 4!$.

Finally, we move on to the last case of when $6$ is the fifth number. We can do a similar casework analysis as we did to find the previous case. We will break up the cases based on which number is placed in the first term.

First, consider if $2$ is the first term. Now, if the second term is $3$, then we need $4$ and $5$ to be both placed in the first block. Otherwise, if $4$ and $5$ are placed in different blocks or they're both in the second block, the conditions won't be satisfied. If the third term is $4$, then $5$ has to be in the fourth term, and we have only $1$ way for this to happen. Now, if the third term is $5$, everything is satisfied, and we have $2!$ ways to arrange them. So if $3$ is the second term, we have $1 + 2!$ ways to arrange them. Now, if the second term is $4$, $5$ has to be placed in the first block, which has only $2$ remaining positions. After placing the $5$, we have $2$ remaining spots overall, so we have $2! \cdot 2$ ways for this to happen. Finally, if $5$ is the second term, everything is satisfied, so we have $3!$ ways to arrange the rest of the numbers. If $2$ is the first term, we have $3! + 2! \cdot 2 + 1 + 2!$ ways.

Now, let's consider $3$ being the first term. We will do this case a little differently. We will consider the position of the number $5$. Let's say $5$ is the second term. Then everything is satisfied, and we have $3!$ ways for this to happen. If $5$ is the third term, everything is once again satisfied, and we have $3!$ ways. If $5$ is the fourth term, we need the number $4$ to be either in the second or third positions, with, after the $4$ is placed, leaving $2$ remaining positions with $2!$ ways to arrange them. Hence, we have $2 \cdot 2!$. If $3$ is the first term, we have $3! + 3! + 2 \cdot 2!$ ways.

Now, if $4$ is the first term, all we need is to satisfy "at least one of the first $4$ terms is greater than $4$." Therefore, $5$ must be placed in the first block where there are $3$ positions. The order of the rest of the numbers doesn't matter. We have $3! \cdot 3$ ways for this to happen.

Finally, if $5$ is the first term, everything is satisfied, and we have $4!$ ways for this to happen.

Finally, the grand total is

$5! + 4! \cdot 4 + 3! \cdot 3 + 4! \cdot 3 + 2 \cdot 2! + 2 \cdot 3! + 2! \cdot 2 \cdot 4 + 2! \cdot 2 + 4! + 4! + 3! + 2 \cdot 2! + 1 + 2! + 3! + 3! + 2 \cdot 2! + 3 \cdot 3! + 4! = \boxed{461}$.

~ilikemath247365