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In a class of $n$ students, each student is given the choice of solving either one of $x$ different algebra problems or one of $y$ different geometry problems. How many different outcomes are possible?

I'll consider a special case where $n = 5.$

Consider $\{A, B, C, D, E\}$. One possible group of algebra problem solvers is $\{A, D, E\}$. Since each student solves one problem we count $3-$lists whose elements stand for an algebra problem. We are counting $3-$lists instead of $3-$sets because each problem is fixed to a student. If there are $x$ algebra problems, then there are $x^3$ algebra problems for every $3-$set of students. So, $x^3$ is the coefficient of $\binom53$. Then, the number of students who solve geometry problems is $2$. For example, $\{B, C\}$. For every $2-$set, there are $y^2$ geometry problems. Also, for every situation like $x^3 \cdot \binom 53$, there are $y^2$ possibilities meaning $y^2$ is the coefficient of $x^3 \cdot \binom 53$. Please, see if that makes sense.

Also, is the sum there to capture all the different values of $k$?

Shaniqua
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    For non-negative integers $x$ and $y$, you have the basics of the idea, and one can then argue that the result is $(x+y)^n$. The exposition could be clearer. Yes, we end up summing over $k$ because the number of students who decide to solve an algebra problem could be any of $0$ up to $n$. – André Nicolas May 12 '15 at 15:13

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