2006 AMC 12A Problems/Problem 25

Problem

How many non- empty subsets of have the following two properties?

No two consecutive integers belong to .

If contains elements, then contains no number less than .

Solution

This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem:

How many ways are there to choose elements from an ordered element set without choosing two consecutive members?

Note that if , the answer will be 0. Otherwise, the elements we choose define boxes into which we can drop the remaining elements, with the caveat that each of the middle boxes must have at least one element. This is equivalent to dropping elements into boxes, where each box is allowed to be empty. And this is equivalent to arranging objects, of which are dividers, which we can do in ways.

Now, looking at our original question, we see that the thing we want to calculate is just

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Another way to think about it that you want to choose numbers out of with no consecutive numbers. Each given each configuration, subtract from the -th element in your subset. This converts your configuration into a configuration with elements where the largest possible element is , with no restriction on consectutive numbers. Since this process is easily reversible, we have a bijection. Without consideration of the second condition, we have:

Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than , which translates to subtracting from the "top" of each binomial coefficient. Now we have, after we cancel all the terms where ,

See also

• 2006 AMC 12A Problems