Question

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?

Answer 1

You want to split up the problem into simple, easily counted tasks.

Answer 2

I'd first assign each group their electrical engineers, then assign them their computer engineers.

Answer 3

so 8! / 6! + 12! / 9!

Answer 4

These are sequential tasks, so you multiply their counts...

Answer 5

8 ways

Answer 6

i don't understand how you got to 8 ways.

Answer 7

495 is the answer I just checked on the calculator

Answer 8

Yes, but how exactly do you get that?

Answer 9

For assigning electrical engineers, I would try: \[
\binom{8}{2}\cdot \binom{6}{2}\cdot \binom{4}{2}\cdot \binom{2}{2}
\]

Answer 10

For assigning computer engineers, I would try: \[

\binom{12}{3}\cdot \binom{9}{3}\cdot \binom{6}{3}\cdot \binom{3}{3}
\]

Answer 11

The total count would be the product of these two.

Answer 12

Thanks.

Answer 13

Do you understand it?

Answer 14

Most of it. Why would you multiply rather than add?

Answer 15

You add when you have an option of doing either task. You multiply when you have two do both tasks.

Answer 16

That would make sense since in this question it's an "and".