Find that probability that exactly $ 4 $ students get candy Question: A teacher gives $ 5 $ candies to $ 8 $ students. She gives each candy to a randomly chosen student, without regard to whether the student has received candy. 
a) What is the probability that exactly $ 4 $ of the students get candy?
Here is my approach, I treat the candies to be identical and the students to be distinguishable: There are total of $ \displaystyle \binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} $ ways to distribute $ 5 $ identical candies to $ 8 $ students. Next, there are $ \displaystyle \binom{8}{4} $ ways to choose $ 4 $ students to receive $ 5 $ candies so that each student among the $ 4 $ chosen must receive at least $ 1 $ candy, yielding a total of $ \displaystyle 4 .\binom{8}{4} $ possibilities. So the answer for part a) is $ \displaystyle \frac{4 .\binom{8}{4}}{\binom{12}{4}} $.

b) What is the probability that $ 5 $ different students get the candy?
I interpret this question as the same as: what is the probability that at least $ 5 $ students get the candy? Again, there are $ \displaystyle \binom{8 + 5 - 1}{5 - 1} = \binom{12}{4} $ ways to distribute $ 5 $ identical candies to $ 8 $ students. Next, I will find the number of possibilities that exactly $ 5, 6, 7, $ and $ 8 $ students get candies, respectively. There are $ \displaystyle \binom{8}{5} $ ways that exactly $ 5 $ students get candies. Note that there is no way more than $ 5 $ students get the candy if each student gets at least $ 1 $ candy. So the answer for part b) is $ \displaystyle \frac{\binom{8}{5}}{\binom{12}{4}} $.
Am I approaching the problem correctly? Any comment is appreciated.

I would look at it this way, starting with b:
b) In order for five students to receive the five sweets, one student must receive one sweet each. The probability of the teacher selecting a different student each time equals:
$$\frac{8}{8} \cdot \frac{7}{8} \cdot \frac{6}{8} \cdot \frac{5}{8} \cdot \frac{4}{8} = \frac{6720}{32768} \approx 0.205$$
a) In order for four students to receive five sweets, one student must receive two sweets and three others must receive one. Let the teacher randomly select a first student. The probability of the teacher selecting this student again in the next turn equals $\frac{1}{8}$. If this happens, the teacher must select three different students afterwards, since one student already got two sweets. If this does not happen (i.e. two different students were selected), there are two options in the third turn: one of the two students who already got a sweet are selected, or a new student is selected. The former case occurs with a probability of $\frac{2}{8}$, in which case two different students must be selected afterwards. If the teacher however selects a third unique student, the probability of the teacher selecting one of them on the fourth term equals $\frac{3}{8}$ (in which case a different student must be selected in the last term). If a new student is however selected, one of the four selected students must be selected again in order to receive two sweets, which happens with a probability of $\frac{4}{8}$. All in all, the probability is thus:
$$\frac{8 \cdot 1 \cdot 7 \cdot 6 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 2 \cdot 6 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 6 \cdot 3 \cdot 5}{8^5} + \frac{8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{8^5} \approx 0.512$$
