2018 AIME II Problems/Problem 11

Problem

Find the number of permutations of $1, 2, 3, 4, 5, 6$ such that for each $k$ with $1$ $\leq$ $k$ $\leq$ $5$ , at least one of the first $k$ terms of the permutation is greater than $k$ .

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 (General Case)

First let us look at the General Case of this kind of Permutation: Consider this kind of Permutation of set $S=\{1,2,...,n\}$ for arbitrary $n \in N$

It is easy to count the total number of the permutation ( $N$ ) of $S$ : $N=n!$ For every $i \in S$ , we can divide $S$ into two subsets: $S_{1\to i}=\{1,2,...i\}; S_{i+1\to n}=\{i+1,i+2,...,n\}$ Define permutation $P$ as the permutation satisfy the condition of this problem. Then according to the condition of this problem, for each $i\in \{1,2,...,n-1\}$ , $P$ is not a permutation of set $S_{1\to i}$ . For each $i\in \{1,2,...,n\}$ , mark the number of permutation $P$ of set $S$ as $P_{k}$ , where $k=i$ , mark the number of permutation $P$ for set $S_{i+1\to n}$ as $P_{i}$ ; then, according to the condition of this problem, the permutation for $S_{i+1\to n}$ is unrestricted, so the number of the unrestricted permutation of $S_{i+1\to n}$ is $(n-i)!$ . As a result, for each $i\in \{1,2,...,n\}$ , the total number of permutation $P$ is $P_{k}=P_{i}(n-i)!$ Notice that according to the condition of this problem, if you sum all $P_{k}$ up, you will get the total number of permutation of $S$ , that is, $N=\sum^{n}_{k=1}{P_{k}}=\sum^{n}_{i=1}{P_{i}(n-i)!}=n!$ Put $n=1,2,3,...,6$ , we will have $P_{1}=1$ $P_{2}=1$ $P_{3}=3$ $P_{4}=13$ $P_{5}=71$ $P_{6}=461$ So the total number of permutations satisify this problem is $P_{6}=\boxed{461}$ .

~Solution by $BladeRunnerAUG$ (Frank FYC)

2018 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2018 AIME II Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

Solution 3 (General Case)