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 1
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?
You want to choose numbers out of with no consecutive numbers. For each configuration, we can 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 consecutive 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 ,
Solution 2
Another way of visualizing the solution above would be to use 's and 's. Denote as the numbers we have chosen and as other numbers. Taking an example, assuming we are picking two numbers, we imagine the shape . This notation forces a number between the two chosen numbers, which blocks the two numbers we picked from being consecutive. Now we consider the orientations with this shape. We have remaining numbers.
We need to find the number of ways to place the remaining 's. We can find this by utilizing stars and bars, with the following marker being placed to represent groups: *| - *|*. Now, we have to place numbers within groups, which is . The same concept can be used for the remaining numbers. The rest of the solution continues as above.
Solution by: Everyoneintexas
Solution 3
We have the same setup as in the previous solution.
Note that if , the answer will be 0. Otherwise, the elements we choose define boxes (which divide the nonconsecutive numbers) into which we can drop the remaining elements, with the caveat that each of the middle boxes must have at least one element (since the numbers are nonconsecutive). 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
Solution 4
Let be the numbers of elements in a subset. First we examine the second condition. No elements less than can be put in a subset of size , therefore the "lowest" element that can go into the subset is , whereas the "highest" element that can go into the subset is . This is a total of or possible elements.
Now we consider the first condition. No consecutive elements are allowed. This means that if an element is put into the subset, both and are no longer possibilities. Assume that all subsets are ordered from least to greatest (we are looking for the number of combinations, so we can order these combinations however we want). Then the first element will be (as a reminder, the lowest possible element in a subset is ), the second element will be at least , and so on. After elements are chosen, we will have skipped elements (these are the elements that were "eliminated" as they were consecutive). Therefore, we ignore exactly elements (if we ignore more or less elements, then changes) Since we must ignore elements, we can simply remove those beforehand. ( possible elements
Now we look for the bounds of . We are looking for non-empty subsets, so . If is too large, there will not be enough non-consecutive terms between and . Specifically, the highest element in a subset using "optimal" selection will be or . If , that means s is too large. Therefore ; solving for yields . Now we know that .
We want to know the number of ways to arrange "balls" into identical "boxes" with at most ball per box, for to . This is equivalent to , or .
See also
2006 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Question |
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