2012 AMC 12B Problems/Problem 18

Problem 18

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?

$\textbf{(A)}\ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ 362,880$

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

$\binom{9}{0} + \binom{9}{1} + \cdots + \binom{9}{9} = 2^9=512 ...... \framebox{B}$

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2012 AMC 12B Problems/Problem 18

Problem 18

Solution