A family of $7$ are out to lunch and each of them orders one dish from a menu of $12$ dishes.
I tried $_{12}C_7$ for both (a) & (b), but I don't understand what's the difference between each perspective. Aren't they the same?
On
The chef doesn't care who orders what - his job is just to cook the meals. However, on the other hand, the waiter needs to give the meal to the right person - in that sense, he does care who ordered what.
In that light, you would find the number of permutations from the waiter's perspective and the number of combinations from the chef's.
Note, however, that you must find these numbers with repetition accounted for. For example, you posit the answer $_{12}C_7$; this is not correct from either perspective, since it assumes choice without repetition. Each of the $7$ people can choose one of $12$ meals, but they could all (hypothetically) order the same meal, you see?
So we note: for $k$ objects taken with replacement/repetition from a group of $n$, we have
$$\begin{align} \text{# of permutations with reptition} &= n^k\\ \text{# of combinations with reptition} &= _{n+k-1}C_k = \binom{n+k-1}{k} = \frac{(n+k-1)!}{k! (n-1)!}\\ \end{align}$$
Hopefully this should be sufficient to guide you to the right answer, neglecting me overlooking something (such as repetition when I first posted this).
For the chef, it’s the number of solutions to $n_1+n_2+ \dots + n_{12} = 7$ ($n_i$ is the number of diners ordering menu item number $i$), or $18\choose7$. For the waiter, it’s $12^7$.