I have the following task:
At a seminar are 3 doctorands and each of them gives 1 presentation. There are 3 master students and each of them gives 2 presentations. There are 3 bachelor students and each of them gives 2 presentations. And there is a professor that gives only one presentation.
How many possibilities are there for the order of the lecturers?
I have modelled the problem like that - each presentation is assigned to a lecturer:
$$\{D_1, D_2, D_3, M_1, M_1, M_2, M_2, M_3, M_3, B_1, B_1, B_2, B_2, B_3, B_3, P\}$$
Now my guess was to get the possibilities for the order for all of the presentations and subtracting all the possibilities, where a lecturer with 2 presentations gives them consecutively.
For the order of presentations, where each lecturer gives only one presentation until starts its own one (one example)
$$presentations:=\{D_1, D_2, D_3, M_1, M_2, M_1, M_3, M_2, M_3, B_1, B_2, B_1, B_3, B_2, B_3, P\}$$
there results the same order of lecturers: $$order of lecturers:=\{D_1, D_2, D_3, M_1, M_2, M_1, M_3, M_2, M_3, B_1, B_2, B_1, B_3, B_2, B_3, P\}$$
If we look at the possibilities, where lecturers can give their presentations consecutively (a person with two presentations will give one and after that the next one)
$$presentations:=\{D_1, D_2, D_3, M_1, M_1, M_2, M_2, M_3, M_3, B_1, B_1, B_2, B_2, B_3, B_3, P\}$$
results a shorter order of lecturers:
$$order of lecturers:=\{D_1, D_2, D_3, M_1, M_2, M_3, B_1, B_2, B_3, P\}$$
So I have first calculated the possibilities for $16$ presentations and then subtracted the possibilities for $10$ (possibilities where lecturers give presentations consecutively).
$\frac{16!}{(16-16)!}- \frac{10!}{(10-10)!}= 20 922 789 888 000 - 3 628 800 = 20 922 786 259 200$
Question: Could that be correct? And a more important one: Is there a better and elegant way to solve that task?