Plaese halp with a worked out probability math question.

Question

anonymous · Answer

A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift.  A quality control consultant is to select 6 of these workers for in-depth interviews.  Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45).

anonymous · Answer

How many selections result in exactly 4 workers coming from the day shift?

anonymous · Answer

So I did $$\frac{ 20! }{ 4!(20-4)! }= 4845$$

anonymous · Answer

but the actual answer is supposedly 1453500, which I think is strange and not make sense at all.

anonymous · Answer

@zepdrix I already worked it out, but I am not sure what's wrong.

anonymous · Answer

I don't understand at all how you could get 1,453,500 from exactly 4 workers coming from the day shift?

anonymous · Answer

Since the question:
How many selections result in all 6 workers coming from the day shift? 
$$\frac{ 20! }{ 6!(20-6)! }=38760$$
is smaller than 1,453,500, IS MY PROFESSOR WRONG?

kropot72 · Answer

The solution is found from the following:
$$\large C(20, 4)\times C(25, 2)=1453500$$

anonymous · Answer

Could you explain that a little more please.

anonymous · Answer

Where does the 25 and 2 come from?

kropot72 · Answer

The number of combinations of the 20 day shift workers taken 4 at a time is C(20, 4). Each of these C(20, 4) combinations can be taken with each of the number of combinations of the remaining 45 - 20 = 25 workers taken 6 - 4 = 2 at a time.

anonymous · Answer

Oh I am so silly thank you!

kropot72 · Answer

You're welcome :)

anonymous · Answer

I just did one part of it :^), you have shed the light in my dark cloud of misunderstanding, I really appreciate it!

kropot72 · Answer

Glad to have been able to clarify the method.