AMC12 2012 B

AMC12 2012 B · Q18

AMC12 2012 B · Q18. It mainly tests Combinations, Casework.

Let $(a_1,a_2, \dots ,a_{10})$ be a list of the first 10 positive integers such that for each $2 \le i \le 10$ either $a_i+1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?

设 $(a_1, a_2, \dots, a_{10})$ 是前 10 个正整数的一个排列，使得对于每个 $2 \le i \le 10$，$a_i + 1$ 或 $a_i -1$ 或两者都在 $a_i$ 之前的位置出现过。有多少这样的排列？

(A) 120 120

(B) 512 512

(D) 181440 181440

(E) 362880 362880

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let $1\leq k\leq 10$. Assume that $a_1=k$. If $k<10$, the first number appear after $k$ that is greater than $k$ must be $k+1$, otherwise if it is any number $x$ larger than $k+1$, there will be neither $x-1$ nor $x+1$ appearing before $x$. Similarly, one can conclude that if $k+1<10$, the first number appear after $k+1$ that is larger than $k+1$ must be $k+2$, and so forth. On the other hand, if $k>1$, the first number appear after $k$ that is less than $k$ must be $k-1$, and then $k-2$, and so forth. To count the number of possibilities when $a_1=k$ is given, we set up $9$ spots after $k$, and assign $k-1$ of them to the numbers less than $k$ and the rest to the numbers greater than $k$. The number of ways in doing so is $9$ choose $k-1$. Therefore, when summing up the cases from $k=1$ to $10$, we get \[\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}\]

令 $1\leq k\leq 10$，并设 $a_1=k$。若 $k<10$，则在 $k$ 之后第一次出现且大于 $k$ 的数必须是 $k+1$；否则若它是任意大于 $k+1$ 的数 $x$，则在 $x$ 之前既不会出现 $x-1$ 也不会出现 $x+1$。同理可得，若 $k+1<10$，则在 $k+1$ 之后第一次出现且大于 $k+1$ 的数必须是 $k+2$，依此类推。另一方面，若 $k>1$，则在 $k$ 之后第一次出现且小于 $k$ 的数必须是 $k-1$，然后是 $k-2$，依此类推。为计数在给定 $a_1=k$ 时的可能性，在 $k$ 之后设置 9 个位置，并将其中 $k-1$ 个位置分配给小于 $k$ 的数，其余位置分配给大于 $k$ 的数。这样做的方式数为 $9$ choose $k-1$。因此对 $k=1$ 到 $10$ 求和，得到 \[\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}\]

Topics