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2012 AMC 10B Problem 22

Problem 22 of 25HarderCounting & Probability

Let (a1,a2,,a10)(a_1,a_2,\ldots,a_{10}) be a list of the first 1010 positive integers such that for each 2i102\le i\le10 either ai+1a_i + 1 or ai1a_i-1 or both appear somewhere before aia_i in the list. How many such lists are there?

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Solution

Every prefix a1,a2,,aia_1,a_2,\ldots,a_i must form an interval of consecutive integers: each new term is required to be one more or one less than a term already present. In particular, the full set is the interval [1,10].[1,10]. Read the list backward. At each step, the term removed must be either the smallest or the largest integer in the current interval; removing an interior term would leave that term with no adjacent predecessor in the forward list. Conversely, every sequence of endpoint removals produces a valid list. There are two choices for each of a10,a9,,a2,a_{10},a_9,\ldots,a_2, after which a1a_1 is forced. Hence there are 29=5122^9=512 lists. Thus, the correct answer is B .

More practice

Concepts: arrangements with restrictions · recursive counting · work backwards

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.