2012 AMC 10B Problems/Problem 22

Problem 22

Let ( $a_1$ , $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ 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

If we have 1 as the first number, then the only possible list is $(1,2,3,4,5,6,7,8,9,10)$ .

If we have 2 as the first number, then we have 9 ways to choose where the one goes, and the numbers ascend from the first number, 2, with the exception of the 1. For example, $(2,3,1,4,5,6,7,8,9,10)$ , or $(2,3,4,1,5,6,7,8,9,10)$ . There are $\dbinom{9}{1}$ ways to do so.

If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are $\dbinom{9}{2}$ ways to do this.

In the same way, the total number of lists is: $\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{10}$

By the binomial theorem, this is $2^{9}$ = $512$ , or $(A)$

                                                                                                                                    Solution 2

Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is $(1,2,3,4,5...10)$ . There are 9 arrows, so the answer is $2^{9}$ = $512$

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2012 AMC 10B Problems/Problem 22

Problem 22