2016 AMC 10A Problems/Problem 20
For some particular value of , when is expanded and like terms are combined, the resulting expression contains exactly terms that include all four variables and , each to some positive power. What is ?
All the desired terms are in the form , where (the part is necessary to make stars and bars work better.) Since , , , and must be at least ( can be ), let , , , and , so . Now, we use stars and bars (also known as ball and urn) to see that there are or solutions to this equation. We notice that , which leads us to guess that is around these numbers. This suspicion proves to be correct, as we see that , giving us our answer of
Note: An alternative is instead of making the transformation, we "give" the variables 1, and then proceed as above.
By Hockey Stick Identity, the number of terms that have all raised to a positive power is . We now want to find some such that . As mentioned above, after noticing that , and some trial and error, we find that , giving us our answer of
Solution 3 (Casework)
The terms are in the form , where . The problem becomes distributing identical balls to different boxes such that each of the boxes has at least ball. The balls in a row have gaps among them. We are going to put or divisors into those gaps. There are cases of how to put the divisors.
Case : Put 4 divisors into gaps. It corresponds to each of has at least one term. There are terms.
Case : Put 3 divisors into gaps. It corresponds to each of has at least one term. There are terms.
So, there are terms. , and since we have
Video Solution by OmegaLearn
Video Solution 2
https://youtu.be/TpG8wlj4eRA with 5 Stars and Bars examples preceding the solution. Time stamps in description to skip straight to solution.
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