What is the probability of getting a specific distribution between students? 
There are 3 different types of exams - A, B, C, each has 7 copies. 15 students are getting a copy randomly. What is the probability that 3 students will get type A exam, 5 students will get type B exam, and 7 students will get type C exam?

Well, obviously the number of different combinations is $\frac{15!}{3!5!7!}$, but I'm not sure how to calculate the probability from here.
 A: The corresponding distribution is called multinomial distribution. It is a generalization of binomial distribution.
$$\frac{15!}{3!5!7!}\left(\frac{1}{3}\right)^{3}\left(\frac{1}{3}\right)^{5}\left(\frac{1}{3}\right)^{7}=\frac{15!}{3!5!7!}\left(\frac{1}{3}\right)^{15}$$
Remark: This is not the right solution. I read the question wrongly.
Andre's solution is correct: 
$$\frac{\binom{7}{3}\binom{7}{5}\binom{7}{7}}{\binom{21}{15}}.$$ 
A: We solve the problem in a couple of ways. The first is slow and plodding, but perhaps more natural. The second is quick.
Slow way: Let us assume that the various exams are labelled, to make them distinct. This will make no difference to the probability.
Imagine giving out  exam papers  to the students, who are lined up in order of student number. Student 1 can be given her exam paper in $21$ ways. For each of these, Student 2 can be given her exam paper in $20$ ways, and so on, for a total of $(21)(20)\cdots(7)$ ways. These are equally likely.
Now we count the favourables, in which the counts are the ones given in the problem.
The $3$ students who get a Type A exam can be chosen in $\binom{15}{3}$ ways. For each of these ways, the actual exam papers can be distributed in $(7)(6)(5)$ ways.
For each of these ways, the students who get a Type B exam can be chosen in $\binom{12}{5}$ ways, and the actual exam papers can be assigned in $(7)(6)(5)(4)(3)$ ways. Finally, the students who get the Type C exam can be chosen in $\binom{7}{7}$ ways, and the exam papers assigned in $7!$ ways.
So the number of favourables is $\binom{15}{3}(7)(6)(5)\binom{12}{5}(7)(6)(5)(4)(3)\binom{7}{7}7!$.
For the probability, divide the number of favourables by $(21)(20)\cdots (7)$.
Quick way: There are $\binom{21}{15}$ equally likely ways to choose the exam copies that will be handed out. There are $\binom{7}{3}\binom{7}{5}\binom{7}{7}$ favourables.
For there are $\binom{7}{3}$ ways to choose which Type A exam papers will be handed out. For each of these ways there are $\binom{7}{5}$ ways to choose which Type B papers will be handed out.
Divide. We get that the required probability is
$$\frac{\binom{7}{3}\binom{7}{5}\binom{7}{7}}{\binom{21}{15}}.$$
