In addition to the algebraic calculation that you asked for, there is a combinatorial explanation of the result. We have a class of $a$ students. We want to pick $b$ lucky students from this class to give a chocolate bar to, and choose $c$ of the lucky students to give an additional chocolate bar to. In how many ways can we do this? Here are two ways of counting:
First count: There are $\binom{a}{b}$ ways to choose the students who will get at least one chocolate bar. For every one of these ways, there are $\binom{b}{c}$ ways to choose the subgroup of $c$ students who will get an additional chocolate bar, for a total of $\binom{a}{b}\binom{b}{c}$.
Second count: Pick the $c$ students who will get $2$ chocolate bars first. There are $\binom{a}{c}$ ways to do this. Now pick an additional $b-c$ students from the remaining $a-c$ to give $1$ chocolate bar to. There are $\binom{a}{c}$ ways to pick the twice lucky students, and for every one of these ways, there are $\binom{a-c}{b-c}$ ways to pick the single chocolate bar students, for a total of $\binom{a}{c}\binom{a-c}{b-c}$ ways.
We counted something in two different ways. So the answers must be the same. But that gives precisely your formula.