Number of ways to distribute $n$ items to $k$ groups given number of items required for each group. The problem goes as follows:

we have 20 workers and we need to distribute them to 4 departments,
department 1 needs 6 workers, department 2 needs 4 workers, department
3 needs 5 workers, department 4 needs 5 workers, by how many ways can
distribute these workers?

My attempt:

since workers are not identical so the order is important then the
answer should be   = $$4^6 \times 3^4 \times 2^5 \times1^5$$

but I do not believe that to be the correct answer as one department can have the same workers but different order and still be considered another way to arrange the workers,

I tried fixing this by dividing by possible arrangments of workers in
each department
= $$(4^6 \times 3^4 \times 2^5 \times1^5) \over 6! \times  4! \times5!  \times 5!$$ $$$$

I ended up having a decimal number so my answer is of course incorrect
what has gone wrong? is my intuition incorrect?
thanks in advance.

NOTE: I know the problem is not the same as what the title says but I
wanted to generalize

 A: Notice that when you do $4^6$ you are essentially taking $6$ workers out of nowhere to place them in one of the $4$ departments. This never guaranties that you have $6$ workers in the first department.

Edit: When you do $4^6$ means that you are going to list tuples of length $6$ where each entry can have one of $4$ values. So you are counting $(d_1,d_2,d_3,d_4,d_5,d_6)$ where $d_i$ is the department in which the $i-$th person is working. Notice that these $6$ people are not specified, so you have to specify them by picking them. Furthermore, when you pick them, nothing guarantees that you will choose the tuple $(4,4,4,4,4,4)$ so in these arrangement you have put all your six workers in the department 4 (but this is not possible because department $4$ has just $5$ people).

The correct approach is, you have your workers, then lets say you want to build department $1,$ then you choose $6$ workers out of the total $20$ in $\binom{20}{6}$ ways. How many workers are left? $20-6,$ now choose the workers for department $2.$ Use the multiplication principle.
You should get something like $$\binom{20}{6}\binom{14}{4}\binom{10}{5}=\frac{20!}{6!\cdot 4!\cdot 5!\cdot 5!}.$$
