Suppose there are 12 red sweets and 12 green sweets. In how many ways can the sweets be distributed so that each of the five children gets at least one sweet of each colour?
My attempt involved splitting this up into two subproblems; one that involves red sweets and one that involves green sweets. For the red sweets, I distributed one sweet to each of the five children first since they "at least" receive one. Then, I use the multiset formula of $${n+k-1\choose{k-1}},$$ for $n=7$ and $k=5$. Evaluating this gives $${11\choose{4}}=330.$$ The answer for the green sweets gives the same. But then, when combining the two; do I square the quantities to get $330^2$ or is it instead $2\cdot 330$?
You will multiply, because once you've picked a way to hand out red sweets you can now independently pick a way to hand out green sweets.
Incidentally, usually one would derive the formula for nonnegative integer solutions to an equation $\sum_{i=1}^n x_i=k$ (i.e. counting multisets) from the formula for positive integer solutions. This is because in the positive integer case you can use "stars and bars" type reasoning. So passing back to the nonnegative case is slightly weird logic. It's still correct, though.