Here's the question you clicked on:
LordHades
There are 8 electrical engineers and 12 computer engineers who are doing a group project. We need to divide them into 4 groups of 5 so that each group contains exactly 2 electrical engineers and 3 computer engineers. How many ways can this be done?
You want to split up the problem into simple, easily counted tasks.
I'd first assign each group their electrical engineers, then assign them their computer engineers.
so 8! / 6! + 12! / 9!
These are sequential tasks, so you multiply their counts...
i don't understand how you got to 8 ways.
495 is the answer I just checked on the calculator
Yes, but how exactly do you get that?
For assigning electrical engineers, I would try: \[ \binom{8}{2}\cdot \binom{6}{2}\cdot \binom{4}{2}\cdot \binom{2}{2} \]
For assigning computer engineers, I would try: \[ \binom{12}{3}\cdot \binom{9}{3}\cdot \binom{6}{3}\cdot \binom{3}{3} \]
The total count would be the product of these two.
Most of it. Why would you multiply rather than add?
You add when you have an option of doing either task. You multiply when you have two do both tasks.
That would make sense since in this question it's an "and".