## anonymous 3 years ago 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?

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

8 ways

6. anonymous

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

7. anonymous

495 is the answer I just checked on the calculator

8. anonymous

Yes, but how exactly do you get that?

9. anonymous

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

10. anonymous

For assigning computer engineers, I would try: $\binom{12}{3}\cdot \binom{9}{3}\cdot \binom{6}{3}\cdot \binom{3}{3}$

11. anonymous

The total count would be the product of these two.

12. anonymous

Thanks.

13. anonymous

Do you understand it?

14. anonymous

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

15. anonymous

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

16. anonymous

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