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2012 AMC 12B Problems/Problem 18

Revision as of 00:31, 1 September 2015 by Daniellionyang (talk | contribs) (→‎Solution 2 (Noticing Stuff))

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 1

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

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

Solution 2 (Noticing Stuff)

If there is only 1 number, the number of ways to order would be 1. If there are 2 numbers, the number of ways to order would be 2. If there are 3 numbers, by listing out, the number of ways is 4. We can then make a conjecture that the problem is simply powers of 2. 

$2^{10-1}=512$.

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See Also

2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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